In the previous section, we completed a discussion on differential equations. In this section, we will see some miscellaneous examples.
Solved example 25.78
For each of the differential equations given below, indicate it's order and degree (if defined).
$\small{\begin{array}{ll}{~\color{magenta} {} } &{{(i)}} &{\frac{d^2y}{dx^2}~+~5x\left(\frac{dy}{dx} \right)^2~-~6y} & {~=~} &{\log x} \\
{~\color{magenta} {} } &{{(ii)}} &{\left(\frac{dy}{dx} \right)^3~-~4 \left(\frac{dy}{dx} \right)^2~+~7y} & {~=~} &{\sin x} \\
{~\color{magenta} {} } &{{(iii)}} &{\frac{d^4y}{dx^4}~-~\sin\left(\frac{d^3 y}{d x^3} \right)} & {~=~} &{0} \\
\end{array}}$
Solution:
Part (i):
• The given differential equation is:
$\small{\frac{d^2y}{dx^2}~+~5x\left(\frac{dy}{dx} \right)^2~-~6y~=~\log x}$
• It is in the polynomial form and hence, degree is defined.
• There are two derivatives: $\small{\frac{d^2y}{dx^2}~~{\rm{and}}~~\frac{dy}{dx}}$
• The highest order derivative is: $\small{\frac{d^2y}{dx^2}}$
♦ It's order is 2.
♦ It's power is 1.
• So for the differential equation:
♦ Order is 2.
♦ Degree is 1.
Part (ii):
• The given differential equation is:
$\small{\left(\frac{dy}{dx} \right)^3~-~4 \left(\frac{dy}{dx} \right)^2~+~7y~=~\sin x}$
• It is in the polynomial form and hence, degree is defined.
• There is only one derivative: $\small{\frac{dy}{dx}}$
♦ It's order is 1.
♦ It's highest power is 3.
• So for the differential equation:
♦ Order is 1.
♦ Degree is 3.
Part (iii):
• The given differential equation is:
$\small{\frac{d^4y}{dx^4}~-~\sin\left(\frac{d^3 y}{d x^3} \right)~=~0}$
• It is not in the polynomial form and hence, degree is not defined.
• There are two derivatives: $\small{\frac{d^4y}{dx^4}~~{\rm{and}}~~\frac{d^3y}{dx^3}}$
• The highest order derivative is: $\small{\frac{d^4y}{dx^4}}$
♦ It's order is 4.
• So for the differential equation:
♦ Order is 4.
Solved example 25.79
For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.
$\small{\begin{array}{ll}{~\color{magenta} {} } &{{(i)}} &{xy=ae^x + be^{-x} + x^2} & {~:~} &{x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2= 0} \\
{~\color{magenta} {} } &{{(ii)}} &{y=e^x(a \cos x + b \sin x)} & {~:~} &{\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0} \\
{~\color{magenta} {} } &{{(iii)}} &{y=x \sin (3x)} & {~:~} &{\frac{d^2y}{dx^2} + 9y - 6 \cos (3x) = 0} \\
{~\color{magenta} {} } &{{(iv)}} &{x^2 = 2y^2 \log y} & {~:~} &{(x^2 + y^2)\frac{dy}{dx} - xy=0} \\
\end{array}}$
Solution:
Part (i)
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{xy} & {~=~} &{ae^x + be^{-x} + x^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{x \frac{dy}{dx}}~+y} & {~=~} &{ae^x -be^{-x} + 2x} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{x \frac{d^2y}{dx^2}}~+~\frac{dy}{dx}~+~\frac{dy}{dx}} & {~=~} &{ae^x +be^{-x} + 2}\\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{x \frac{d^2y}{dx^2}}~+~2 \frac{dy}{dx}} & {~=~} &{ae^x +be^{-x} + 2}\\
\end{array}}$
2. Substitute the above derivatives in the given differential equation:
• The given differential equation is:
$\small{x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2= 0}$
• Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left[x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} \right] - \left[xy \right] + x^2 - 2} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left[ae^x +be^{-x} + 2 \right] - \left[ae^x + be^{-x} + x^2 \right] + x^2 - 2} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{ae^x +be^{-x} + 2 - ae^x - be^{-x} - x^2 + x^2 - 2} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{ 2 - x^2 + x^2 - 2} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{0} & {~=~} &{0} \\
\end{array}}$
3. L.H.S is equal to R.H.S
• So the given function is indeed a solution of the corresponding differential equation.
Part (ii)
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{e^x(a \cos x + b \sin x)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{y} & {~=~} &{a e^x \cos x + b e^x \sin x} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{a\left[e^x(-\sin x)+(\cos x)e^x \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~b\left[e^x(\cos x)+(\sin x)e^x \right]} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{-ae^x \sin x~+~a e^x \cos x} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~b e^x \cos x~+~b e^x \sin x} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{-a\left[e^x(\cos x)+(\sin x)e^x \right]~+~a \left[e^x(-\sin x)+(\cos x)e^x \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~b\left[e^x(-\sin x)+(\cos x)e^x \right]~+~b\left[e^x(\cos x)+(\sin x)e^x \right]} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{-ae^x(\cos x)-a e^x(\sin x) ~-~a e^x(\sin x)+a e^x(\cos x) } \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{-~be^x(\sin x)+be^x(\cos x) ~+~be^x(\cos x)+b e^x(\sin x) } \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{-a e^x(\sin x) ~-~a e^x(\sin x) } \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+be^x(\cos x) ~+~be^x(\cos x) } \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{-2a e^x(\sin x) } \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+2be^x(\cos x) } \\
\end{array}}$
2. Substitute the above derivatives in the given differential equation:
• The given differential equation is:
$\small{x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2= 0}$
Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left[\frac{d^2y}{dx^2} \right] - \left[2\frac{dy}{dx} \right] + \left[2y \right] } & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left[-2a e^x(\sin x)+2be^x(\cos x) \right]} & {{}} &{{}} \\
{~\color{magenta} {} } &{{}} &{-\left[-2ae^x \sin x~+~2a e^x \cos x +~2b e^x \cos x~+~2b e^x \sin x \right]} & {{}} &{{}} \\
{~\color{magenta} {} } &{{}} &{+\left[2a e^x \cos x + 2b e^x \sin x \right]} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{-2a e^x(\sin x)+2be^x(\cos x) } & {{}} &{{}} \\
{~\color{magenta} {} } &{{}} &{+2ae^x \sin x~-~2a e^x \cos x -~2b e^x \cos x~-~2b e^x \sin x } & {{}} &{{}} \\
{~\color{magenta} {} } &{{}} &{+2a e^x \cos x + 2b e^x \sin x } & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{0} & {~=~} &{0} \\
\end{array}}$
3. L.H.S is equal to R.H.S
• So the given function is indeed a solution of the corresponding differential equation.
Part (iii)
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{x \sin (3x)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{ \frac{dy}{dx}}} & {~=~} &{3x\cos(3x)~+~\sin(3x)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{ \frac{d^2y}{dx^2}}} & {~=~} &{6 \cos\left(3x\right) - 9x \sin\left(3x\right)}\\
\end{array}}$
2. Substitute the above derivatives in the given differential equation:
• The given differential equation is:
$\small{x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2= 0}$
• Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left[\frac{d^2y}{dx^2} \right] + \left[9y \right] - 6 \cos (3x)} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left[6 \cos\left(3x\right) - 9x \sin\left(3x\right) \right] + \left[9(x \sin (3x)) \right] - 6 \cos (3x)} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{6 \cos\left(3x\right) - 9x \sin\left(3x\right) + 9x \sin (3x) - 6 \cos (3x)} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{0} & {~=~} &{0} \\
\end{array}}$
3. L.H.S is equal to R.H.S
• So the given function is indeed a solution of the corresponding differential equation.
Part (iv)
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2} & {~=~} &{2y^2 \log y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{ \frac{dy}{dx}}} & {~=~} &{\frac{x}{2y \log\left(y\right) + y}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{ \frac{dy}{dx}}} & {~=~} &{\frac{x}{\frac{x^2}{y} + y}~=~\frac{xy}{x^2~+~y^2}} \\
\end{array}}$
◼ Remarks:
[(2) magenta color]: The reader should write all the steps related to this implicit differentiation.
2. Substitute the above derivative in the given differential equation:
• The given differential equation is:
$\small{x \frac{d^2y}{dx^2} + 2\frac{dy}{dx} - xy + x^2 - 2= 0}$
• Substituting, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x^2 + y^2)\frac{dy}{dx} - xy} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(x^2 + y^2)\left[\frac{xy}{x^2~+~y^2} \right] - \frac{2y^3 \log y}{x}} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{xy - \frac{2y^3 \log y}{x}} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x - \frac{2y^2 \log y}{x}} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{x^2~-~2y^2 \log y} & {~=~} &{0} \\
\end{array}}$
◼ Remarks:
[(2) magenta color]: Here we replace xy based on the given equation.
3. Based on the given equation, the above L.H.S is equal to R.H.S
• So the given function is indeed a solution of the corresponding differential equation.
Solved example 25.80
Form the differential equation representing the family of curves given by $\small{(x-a)^2~+~2y^2~=~a^2}$, where 'a' is an arbitrary constant.
Solution:
1. To eliminate 'a', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-a)^2~+~2y^2} & {~=~} &{a^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{ \frac{dy}{dx}}} & {~=~} &{\frac{a-x}{2y}} \\
\end{array}}$
2. The remaining 'a' in the above R.H.S can be eliminated from the original equation:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-a)^2~+~2y^2} & {~=~} &{a^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x^2 - 2ax + a^2 + 2y^2} & {~=~} &{a^2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x^2 - 2ax + 2y^2} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{2ax} & {~=~} &{x^2 + 2y^2} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{a} & {~=~} &{\frac{x^2 + 2y^2}{2x}} \\
\end{array}}$
3. Substituting in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{a-x}{2y}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{\frac{x^2 + 2y^2}{2x}-x}{2y}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{2y^2~-~x^2}{4xy}} \\
\end{array}}$
Solved example 25.81
Prove that $\small{x^2~-~y^2~=~c(x^2~+~y^2)^2}$ is the general solution of differential
equation $\small{\left(x^3~-~3x y^2 \right)dx~=~(y^3~-~3x^2 y)dy}$ where 'c' is a parameter.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{x^3 - 3xy^2}{y^3 - 3x^2 y}}$
2. Let $\small{F(x,y)~=~\frac{x^3 - 3xy^2}{y^3 - 3x^2 y}}$
• We need to show that, this is a homogeneous function of degree zero.
(i) We can rearrange F(x,y) as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F(x,y)} & {~=~} &{\frac{x^3 - 3xy^2}{y^3 - 3x^2 y}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{F(x,y)} & {~=~} &{\left[\frac{x^3}{x^3} \right]\left[\frac{1 ~+~ 3\left(\frac{y}{x} \right)^2}{\left(\frac{y}{x} \right)^3 ~-~ 3\left(\frac{y}{x} \right)} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\left[1 \right]\left[\frac{1 ~+~ 3\left(\frac{y}{x} \right)^2}{\left(\frac{y}{x} \right)^3 ~-~ 3\left(\frac{y}{x} \right)} \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{x^0\left[\frac{1 ~+~ 3\left(\frac{y}{x} \right)^2}{\left(\frac{y}{x} \right)^3 ~-~ 3\left(\frac{y}{x} \right)} \right]} \\
{~\color{magenta} 5 } &{} &{} & {~=~} &{x^0\left[g\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\small{g\left(\frac{y}{x} \right)~=~\frac{1 ~+~ 3\left(\frac{y}{x} \right)^2}{\left(\frac{y}{x} \right)^3 ~-~ 3\left(\frac{y}{x} \right)}}$
(ii) We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y) = x^n[g\left(\frac{y}{x}\right)]}$
♦ Where n is a natural number.
• So it is a homogeneous function.
• For this function, n = 0. So degree is zero.
• Therefore, the given differential equation is homogeneous.
Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{y}{x}= v}$
Which is same as: $\small{y = vx}$
2. Differentiating the above equation with respect to x, we get:
$\small{\frac{dy}{dx}~=~v~+~x \frac{dv}{dx}}$
3. Substituting (1) and (2) in the given differential equation, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{1 ~+~ 3\left(\frac{y}{x} \right)^2}{\left(\frac{y}{x} \right)^3 ~-~ 3\left(\frac{y}{x} \right)}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{v~+~x \frac{dv}{dx}} & {~=~} &{\frac{1 ~+~ 3\left(v \right)^2}{\left(v \right)^3 ~-~ 3\left(v \right)}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{1 ~+~ 3\left(v \right)^2}{\left(v \right)^3 ~-~ 3\left(v \right)}~-~v~=~\frac{1 + 3v^2 - v^4 + 3v^2}{v^3-3v}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{1 - v^4}{v^3-3v}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\left[\frac{v^3-3v}{1-v^4} \right]dv} & {~=~} &{\left[\frac{1}{x} \right]dx} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[\frac{v^3-3v}{1-v^4} \right]dv}} & {~=~} &{\int{\left[\frac{1}{x} \right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{\int{\left[\frac{v^3}{1-v^4} \right]dv}~-~\int{\left[\frac{3v}{1-v^4} \right]dv}} & {~=~} &{\int{\left[\frac{1}{x} \right]dx}} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{-\frac{\log\left(v^{4} - 1\right)}{4}~-~\left[\frac{3 \left(\ln\left(v^{2} + 1\right) - \ln\left(v^{2} - 1\right)\right)}{4} \right]~+~\rm{C}_1} & {~=~}&{\log \left|x \right|~+~\rm{C}_2} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{\frac{3\log\left( v^2 - 1 \right)~-~3\log\left( v^2 + 1 \right)~-~\log\left( v^4 - 1 \right)}{4}} & {~=~}&{\log \left|x \right|~+~\rm{C}_3} \\
{~\color{magenta} 5 }&{{\Rightarrow}} &{\log\left[\frac{(v^2 - 1)^{3/4}}{(v^2 + 1)^{3/4}~\times~(v^4 - 1)^{1/4}} \right]} & {~=~}&{\log \left|x \right|~+~\rm{C}_3} \\
{~\color{magenta} 6 }&{{\Rightarrow}} &{\log\left[\frac{(v^2 - 1)^{3/4}}{(v^2 + 1)^{3/4}~\times~(v^2 - 1)^{1/4}~\times~(v^2 + 1)^{1/4}} \right]} & {~=~}&{\log \left|x \right|~+~\rm{C}_3} \\
{~\color{magenta} 7 }&{{\Rightarrow}} &{\log\left[\frac{(v^2 - 1)^{2/4}}{v^2 + 1} \right]} & {~=~}&{\log \left|x \right|~+~\rm{C}_3} \\
{~\color{magenta} 8 }&{{\Rightarrow}} &{\log\left[\frac{(v^2 - 1)^{2/4}}{(v^2 + 1)x} \right]} & {~=~}&{\log\left[\rm{C}_4 \right]} \\
{~\color{magenta} 9 }&{{\Rightarrow}} &{\frac{(v^2 - 1)^{2/4}}{(v^2 + 1)x} } & {~=~}&{\rm{C}_4 } \\
\end{array}}$
• The reader may write all steps related to the integration in [(2) magenta color]
5. Replacing 'v', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{(v^2 - 1)^{1/2}}{(v^2 + 1)x}} & {~=~} &{\rm{C}_4} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{\frac{(y^2 - x^2)^{1/2}\,x^2}{(y^2 + x^2)x^2}} & {~=~} &{\rm{C}_4} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{\frac{y^2 - x^2}{(y^2 + x^2)^2}} & {~=~} &{\left[\rm{C}_4 \right]^2} \\
{~\color{magenta} 4 }&{{\Rightarrow}}&{\frac{x^2 - y^2}{(y^2 + x^2)^2}} & {~=~} &{(-1)\left[\rm{C}_4 \right]^2} \\
{~\color{magenta} 5 }&{{\Rightarrow}}&{\frac{x^2 - y^2}{(y^2 + x^2)^2}} & {~=~} &{\rm{C}} \\
{~\color{magenta} 6 }&{{\Rightarrow}}&{x^2 - y^2} & {~=~} &{{\rm{C}}(x^2 + y^2)^2} \\
\end{array}}$
Solved example 25.82
Form the differential
equation of the family of circles in the first quadrant which touch the coordinate axes.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x−a)^2 ~+~ (y−b)^2 ~=~ r^2}$
•
The family mentioned in the question will be similar to the one shown in fig.25.18 below:
 |
| Fig.25.18 |
• Based on the fig., we can write: a = b = r
• So the equation of the family becomes:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-r)^2~+~(y-r)^2} & {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x^2~+~y^2~-~2xr~-~2yr~+~r^2} & {~=~} &{0} \\
\end{array}}$
• In the above fig.25.18,
♦ r = 1 for the green circle
♦ r = 2 for the yellow circle
♦ r = 3 for the red circle
2. To eliminate 'r', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~+~y^2~-~2xr~-~2yr~+~r^2} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2x~+~2y\,y'~-~2r~-~2r\,y'} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x~+~y\,y'~-~r~-~r\,y'} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x~+~y\,y'~-~r(1 + y')} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{r} & {~=~} &{\frac{x + y y'}{1 + y'}} \\
\end{array}}$
• Next we substitute this value of r in the original equation of the circle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-r)^2~+~(y-r)^2} & {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left[x~-~\frac{x + y y'}{1 + y'}\right]^2~+~\left[y~-~\frac{x + y y'}{1 + y'}\right]^2} & {~=~} &{\left[\frac{x + y y'}{1 + y'} \right]^2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\left[x + x y'~-~x - yy'\right]^2~+~\left[y + yy'- x - yy'\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\left[(x-y)y'\right]^2~+~\left[y-x\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\left[(x-y)y'\right]^2~+~\left[x-y\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{(x-y)^2(y')^2~+~(x-y)^2} & {~=~} &{(x + y y')^2} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{(x-y)^2\,\left[(y')^2~+~1 \right]} & {~=~} &{(x + y y')^2} \\
\end{array}}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation:
$\small{(x-y)^2\,\left[(y')^2~+~1 \right]~=~(x + y y')^2}$ represents the given family of circles.
Solved example 25.83
Form the differential
equation of the family of circles in the second quadrant which touch the coordinate axes.
Solution:
1. We know that, the circle with center at (a,b) and radius r is given by the equation: $\small{(x−a)^2 ~+~ (y−b)^2 ~=~ r^2}$
•
The family mentioned in the question will be similar to the one shown in fig.25.19 below:
 |
| Fig.25.19 |
• Based on the fig., we can write: a = b = r
• So the equation of the family becomes:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{(x+r)^2~+~(y-r)^2}
& {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x^2~+~y^2~+~2xr~-~2yr~+~r^2} & {~=~} &{0} \\
\end{array}}$
• In the above fig.25.18,
♦ r = 1 for the green circle
♦ r = 2 for the yellow circle
♦ r = 3 for the red circle
2. To eliminate 'r', we differentiate the equation as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x^2~+~y^2~+~2xr~-~2yr~+~r^2} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2x~+~2y\,y'~+~2r~-~2r\,y'} & {~=~} &{0} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x~+~y\,y'~+~r~-~r\,y'} & {~=~} &{0} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x~+~y\,y'~+~r(1 - y')} & {~=~} &{0} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{r} & {~=~} &{\frac{x + y y'}{y' - 1}} \\
\end{array}}$
• Next we substitute this value of r in the original equation of the circle:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x+r)^2~+~(y-r)^2} & {~=~} &{r^2} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left[x~+~\frac{x + y y'}{y' - 1}\right]^2~+~\left[y~-~\frac{x + y y'}{y' - 1}\right]^2} & {~=~} &{\left[\frac{x + y y'}{y' - 1} \right]^2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\left[x y'~-~x~+~x + yy'\right]^2~+~\left[yy' - y - x - yy'\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\left[(x+y)y'\right]^2~+~\left[-y-x\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\left[(x+y)y'\right]^2~+~\left[(-1)(x+y)\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\left[(x+y)y'\right]^2~+~\left[x+y\right]^2} & {~=~} &{\left[x + y y' \right]^2} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{(x+y)^2(y')^2~+~(x+y)^2} & {~=~} &{(x + y y')^2} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{(x+y)^2\,\left[(y')^2~+~1 \right]} & {~=~} &{(x + y y')^2} \\
\end{array}}$
3. We got a differential equation without arbitrary constants.
• We can write:
the differential equation:
$\small{(x+y)^2\,\left[(y')^2~+~1 \right]~=~(x + y y')^2}$ represents the given family of circles.
Solved example 25.84
Find the general solution of the differential
equation $\small{\frac{dy}{dx}~+~\sqrt{\frac{1 - y^2}{1 - x^2}}~=~0}$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}~+~\sqrt{\frac{1 - y^2}{1 - x^2}}} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{(-1)\sqrt{\frac{1 - y^2}{1 - x^2}}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{(-1)\sqrt{1-y^2}}{\sqrt{1 - x^2}}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dy}{\sqrt{1-y^2}}} & {~=~} &{\frac{(-1)dx}{\sqrt{1 - x^2}}} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1}{\sqrt{1-y^2}}\right]dy}} & {~=~} &{\int{\left[\frac{(-1)}{\sqrt{1 - x^2}} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\sin^{-1}y~+~\rm{C}_1} & {~=~} &{(-1)\sin^{-1}x~-~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\sin^{-1}x~+~\sin^{-1}y} & {~=~} &{~-~\rm{C}_1~-~\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\sin^{-1}x~+~\sin^{-1}y} & {~=~} &{\rm{C}} \\
\end{array}}$
3. So the general solution is:
$\small{\sin^{-1}x~+~\sin^{-1}y~=~\rm{C}}$
Solved example 25.85
Show that the general solution of the differential
equation $\small{\frac{dy}{dx}~+~\frac{y^2 + y + 1}{x^2 + x + 1}~=~0}$ is given by $\small{x + y + 1~=~A(1 - x - y - 2xy)}$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}~+~\frac{y^2 + y + 1}{x^2 + x + 1}} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{(-1)\frac{y^2 + y + 1}{x^2 + x + 1}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{y^2 + y + 1}} & {~=~} &{\frac{(-1)dx}{x^2 + x + 1}} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1}{y^2 + y + 1}\right]dy}} & {~=~} &{\int{\left[\frac{(-1)}{x^2 + x + 1} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{2 \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right)}{\sqrt{3}}~+~\rm{C}_1} & {~=~} &{(-1)\frac{2 \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)}{\sqrt{3}}~-~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{2 \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right)~+~\sqrt{3}\,\rm{C}_1} & {~=~} &{(-1)2 \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)~-~\sqrt{3}\,\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right)~+~\frac{\sqrt{3}}{2}\,\rm{C}_1} & {~=~} &{(-1) \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)~-~\frac{\sqrt{3}}{2}\,\rm{C}_2} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{2x + 1}{\sqrt{3}}\right)~+~\tan^{-1}\left(\frac{2y + 1}{\sqrt{3}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\frac{2x + 1}{\sqrt{3}}~+~\frac{2y + 1}{\sqrt{3}}}{1~-~\frac{2x + 1}{\sqrt{3}}~\times~\frac{2y + 1}{\sqrt{3}}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\frac{2x + 1 + 2y + 1}{\sqrt{3}}}{\frac{3 - (2x + 1)(2y+1)}{3}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\frac{2x + 2y + 2}{\sqrt{3}}}{\frac{3 -(4xy + 2x + 2y+ 1)}{3}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 9 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\frac{2x + 2y + 2}{\sqrt{3}}}{\frac{2 -4xy - 2x - 2y}{3}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 10 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\frac{x + y + 1}{\sqrt{3}}}{\frac{1 -2xy - x - y}{3}}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 11 } &{{\Rightarrow}} &{ \tan^{-1}\left(\frac{\sqrt{3}(x+y+1)}{1 -2xy - x - y}\right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 12 } &{{\Rightarrow}} &{ \frac{\sqrt{3}(x+y+1)}{1 -2xy - x - y}} & {~=~} &{\tan \left[\rm{C}_3 \right]} \\
{~\color{magenta} 13 } &{{\Rightarrow}} &{ \frac{x+y+1}{1 -2xy - x - y}} & {~=~} &{\frac{\tan \left[\rm{C}_3 \right]}{\sqrt{3}}} \\
{~\color{magenta} 14 } &{{\Rightarrow}} &{ \frac{x+y+1}{1 -2xy - x - y}} & {~=~} &{\rm{A}} \\
{~\color{magenta} 15 } &{{\Rightarrow}} &{ x + y + 1} & {~=~} &{{\rm{A}}(1 - x - y - 2xy)} \\
\end{array}}$
• The reader may write all steps related to the integration in [(1) magenta color]
◼ Remarks:
[(7) magenta color]: Here we use the identity:
$\small{\tan^{-1}a~+~\tan^{-1} b~=~\frac{a + b}{1~-~ab}}$
3. So the general solution is:
$\small{x + y + 1~=~{\rm{A}}(1 - x - y - 2xy)}$
Solved example 25.86
Find the equation of the curve passing through the point $\small{\left(0,\frac{\pi}{4} \right)}$ whose differential
equation is $\small{\sin x \cos y\,dx~+~\cos x \sin y\,dy~=~0}$.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\sin x \cos y\,dx~+~\cos x \sin y\,dy} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\cos x \sin y\,dy} & {~=~} &{(-1)\sin x \cos y\,dx} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{\sin y\,dy}{\cos y}} & {~=~} &{\frac{(-1)\sin x\,dx}{\cos x}} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{\sin y}{\cos y}\right]dy}} & {~=~} &{\int{\left[\frac{(-1)\sin x}{\cos x} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(-1)\log\left(\left|\cos y \right| \right)~+~\rm{C}_1} & {~=~} &{(-1)(-1)\log\left( \left|\cos x \right| \right)~-~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\log\left( \left|\cos x \right| \right)~+~\log\left(\left|\cos y \right| \right)} & {~=~} &{\rm{C}_1~+~\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\log\left( \left|\cos x\,\cos y \right| \right)} & {~=~} &{\rm{C}_3~=~\log\left( \left|\rm{C} \right| \right)} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\cos x\,\cos y} & {~=~} &{\rm{C}} \\
\end{array}}$
3. So the general solution is:
$\small{\cos x\,\cos y~=~\rm{C}}$
4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes π/4 when x = 0
• So substituting x = 0 and y = π/4 in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\cos x\,\cos y} & {~=~}&{\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\cos 0~\times~\cos \left(\frac{\pi}{4} \right)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{1~\times~\frac{1}{\sqrt 2}} & {~=~}&{\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{\frac{1}{\sqrt 2}} \\
\end{array}}$
5. So the particular solution is:
$\small{\cos x\,\cos y~=~\frac{1}{\sqrt 2}}$
$\small{\Rightarrow \cos y~=~\frac{\sec x}{\sqrt 2}}$
Solved example 25.87
Find the particular solution of the differential
equation $\small{\left(1 + e^{2x} \right)dy~+~\left(1 + y^2 \right)e^x\,dx~=~0}$, given that y = 1 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\left(1 + e^{2x} \right)dy~+~\left(1 + y^2 \right)e^x\,dx} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left(1 + e^{2x} \right)dy} & {~=~} &{(-1)\left(1 + y^2 \right)e^x\,dx} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{1 + y^2}} & {~=~} &{\frac{(-1)e^x\,dx}{1 + e^{2x}}} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1}{1 + y^2}\right]dy}} & {~=~} &{\int{\left[\frac{(-1)e^x}{1 + e^{2x}} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\tan^{-1}y~+~\rm{C}_1} & {~=~} &{(-1)\tan^{-1}(e^x)~-~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\tan^{-1}y~+~\tan^{-1}(e^x)} & {~=~} &{~-~\rm{C}_1~-~\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\tan^{-1}y~+~\tan^{-1}(e^x)} & {~=~} &{\rm{C}} \\
\end{array}}$
• The reader may write all steps related to the integration in [(1) magenta color]
3. So the general solution is:
$\small{\tan^{-1}y~+~\tan^{-1}(e^x)~=~\rm{C}}$
4. The constant C can take infinite number of values. So there are infinite number of solutions possible.
• We want that particular solution in which y becomes 1 when x = 0
• So substituting x = 0 and y = 1 in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\tan^{-1}y~+~\tan^{-1}(e^x)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\tan^{-1}(1)~+~\tan^{-1}(e^0)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\tan^{-1}(1)~+~\tan^{-1}(1)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{\pi}{4}~+~\frac{\pi}{4}} & {~=~}&{\rm{C}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{\frac{\pi}{2}} \\
\end{array}}$
5. So the particular solution is:
$\small{\tan^{-1}y~+~\tan^{-1}(e^x)~=~\frac{\pi}{2}}$
Solved example 25.88
Solve the differential
equation $\small{y\,e^{\frac{x}{y}}dx~=~\left(x\,e^{\frac{x}{y}}~+~y^2 \right)dy~(y \ne 0)}$, given that y = 1 when x = 0.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dx}{dy}~=~\frac{x\,e^{\frac{x}{y}}~+~y^2}{y\,e^{\frac{x}{y}}}}$
2. Let $\small{F(x,y)~=~\frac{x\,e^{\frac{x}{y}}~+~y^2}{y\,e^{\frac{x}{y}}}}$
• We need to show that, this is a homogeneous function of degree zero.
(i) We can rearrange F(x,y) as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F(x,y)} & {~=~} &{\frac{x\,e^{\frac{x}{y}}~+~y^2}{y\,e^{\frac{x}{y}}}} \\
{~\color{magenta} 2 } &{\Rightarrow} &{F(x,y)} & {~=~} &{\left[\frac{y^2}{y^2} \right]\left[\frac{\left(\frac{1}{y} \right)\left(\frac{x}{y} \right)e^{\frac{x}{y}}~+~1}{\left(\frac{1}{y} \right)\,e^{\frac{x}{y}}} \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\left[1 \right]\left[\frac{\left(\frac{1}{y} \right)\left(\frac{x}{y} \right)e^{\frac{x}{y}}~+~1}{\left(\frac{1}{y} \right)\,e^{\frac{x}{y}}} \right]} \\
\end{array}}$
Here, $\small{h\left(\frac{y}{x} \right)~\ne~\frac{\left(\frac{1}{y} \right)\left(\frac{x}{y} \right)e^{\frac{x}{y}}~+~1}{\left(\frac{1}{y} \right)\,e^{\frac{x}{y}}}}$
(ii) We see that:
For the function $\small{F(x,y)}$, it is not possible to write:
$\small{F(x,y) = x^n[h\left(\frac{y}{x}\right)]}$
♦ Where n is a natural number.
• So it is not a homogeneous function.
In the next section, we will see a few more miscellaneous examples.
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