In the previous section, we saw some miscellaneous examples on differential equations. In this section, we will see a few more miscellaneous examples.
Solved example 25.89
Find a particular solution of the differential
equation $(x-y)(dx+dy)~=~dx - dy}$, given that y = −1 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x-y)(dx+dy)} & {~=~} &{dx - dy} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x \,dx + x\,dy - y\,dx - y\,dy} & {~=~} &{dx - dy} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{ x\,dy - y\,dy + dy} & {~=~} &{-x \,dx + y\,dx+ dx } \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{( x - y + 1)dy} & {~=~} &{(-x + y+ 1)dx } \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{\frac{-x + y+ 1}{x - y + 1}} \\
\end{array}}$
2. Put $\small{x-y~=~t}$
Differentiating w.r.t x, we get:
Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{1 - \frac{dy}{dx}} & {~=~} &{\frac{dt}{dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{1 - \frac{dt}{dx}} \\
\end{array}}$
• Substituting this in the result in (1), we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{-x + y+ 1}{x - y + 1}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{1 - \frac{dt}{dx}} & {~=~} &{\frac{-t+ 1}{t + 1}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dt}{dx}} & {~=~} &{1~-~\left(\frac{-t+ 1}{t + 1} \right)} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dt}{dx}} & {~=~} &{\frac{t + 1+t- 1}{t + 1}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\frac{dt}{dx}} & {~=~} &{\frac{2t}{t + 1}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\frac{(t+1)dt}{2t}} & {~=~} &{dx} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{\left[\frac{1}{2}~+~\frac{1}{2t} \right]dt} & {~=~} &{\left[1 \right]dx} \\
\end{array}}$
3. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[\frac{1}{2}~+~\frac{1}{2t} \right]dt}} & {~=~} &{\int{\left[1 \right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{\int{\left[\frac{1}{2} \right]dt}~+~\int{\left[\frac{1}{2t} \right]dt}} & {~=~} &{\int{\left[1 \right]dx}} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{\frac{t}{2}~+~\frac{\log\left( \left|t \right| \right)}{2} ~+~\rm{C}_1} & {~=~}&{x~+~\rm{C}_2} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{t~+~\log\left( \left|t \right| \right) ~+~2\rm{C}_1} & {~=~}&{2x~+~2\rm{C}_2} \\
{~\color{magenta} 5 }&{{\Rightarrow}} &{t~+~\log\left( \left|t \right| \right) ~-~2x} & {~=~}&{\rm{C}_3} \\
\end{array}}$
• The reader may write all steps related to the integration in [(2) magenta color]
4. Replacing 't', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{t~+~\log\left( \left|t \right| \right) ~-~2x} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{x-y~+~\log\left( \left|x-y \right| \right) ~-~2x} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{-x-y~+~\log\left( \left|x-y \right| \right)} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 4 }&{{\Rightarrow}}&{\log\left( \left|x-y \right| \right)} & {~=~} &{x+y~+~\rm{C}} \\
\end{array}}$
5. 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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\log\left( \left|x-y \right| \right)} & {~=~}&{x + y + \rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\log\left( \left|0-(-1) \right| \right)} & {~=~}&{0 + (-1) + \rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\log\left( \left|1 \right| \right)} & {~=~}&{-1 + \rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{0} & {~=~}&{-1 + \rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{1} \\
\end{array}}$
6. So the particular solution is:
$\small{\log\left( \left|x-y \right| \right)~=~x + y + \rm{C}}$
Solved example 25.90
Find a particular solution of the differential
equation $(x+1)\frac{dy}{dx}~=~2 e^{-y}~-~1}$, given that y = 0 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x+1)\frac{dy}{dx}} & {~=~} &{2 e^{-y}~-~1} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{2 e^{-y}~-~1}} & {~=~} &{\frac{dx}{x+1}} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1}{2 e^{-y}~-~1}\right]dy}} & {~=~} &{\int{\left[\frac{1}{x+1} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(-1)\log\left(\left|\mathrm{e}^{y} - 2\right|\right)~-~\rm{C}_1} & {~=~} &{\log\left(\left|x + 1\right|\right)~+~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\log\left(\left|x + 1\right|\right)~+~\log\left(\left|\mathrm{e}^{y} - 2\right|\right)} & {~=~} &{~-~\rm{C}_1~-~\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\log\left[(x+1)\left({e}^{y} - 2 \right) \right]} & {~=~} &{\rm{C}_3} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\log\left[(x+1)\left({e}^{y} - 2 \right) \right]} & {~=~} &{\log\rm{C}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{(x+1)\left({e}^{y} - 2 \right)} & {~=~} &{\rm{C}} \\
\end{array}}$
3. So the general solution is:
$\small{(x+1)\left({e}^{y} - 2 \right)~=~\rm{C}}$
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 0 when x = 0
• So substituting x = 0 and y = 0 in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x+1)\left({e}^{y} - 2 \right)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(0+1)\left({e}^{0} - 2 \right)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{(1)\left(1 - 2 \right)} & {~=~}&{\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{-1} \\
\end{array}}$
5. So the particular solution is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x+1)\left({e}^{y} - 2 \right)} & {~=~}&{-1} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{(x+1)\left(2 - {e}^{y} \right)} & {~=~}&{1} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{2 - e^y} & {~=~}&{\frac{1}{x+1}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{e^y} & {~=~}&{2~-~\frac{1}{x+1}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{e^y} & {~=~}&{\frac{2x + 2 - 1}{x+1}~=~\frac{2x + 1}{x+1}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{y\,\log e} & {~=~}&{\log\left( \left|\frac{2x + 1}{x+1} \right| \right)} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{y} & {~=~}&{\log\left( \left|\frac{2x + 1}{x+1} \right| \right)~(x~\ne~-1)} \\
\end{array}}$
Solved example 25.91
Find a particular solution of the differential
equation $\log\left(\frac{dy}{dx} \right)~=~3x + 4y}$, given that y = 0 when x = 0.
Solution:
1. The variables can be separated and the differential equation can be rewritten:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\log\left(\frac{dy}{dx} \right)} & {~=~} &{3x + 4y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{e^{(3x + 4y)}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{dy}{dx}} & {~=~} &{e^{(3x)}~\times~e^{(4y)}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dy}{e^{(4y)}}} & {~=~} &{e^{(3x)}\,dx} \\
\end{array}}$
2. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\int{\left[\frac{1}{e^{(4y)}}\right]dy}} & {~=~} &{\int{\left[e^{(3x)} \right]dx}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{-\frac{\mathrm{e}^{-4y}}{4}~+~\rm{C}_1} & {~=~} &{\frac{e^{3x}}{3}~+~\rm{C}_2} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}} & {~=~} &{\rm{C}_1~-~\rm{C}_2} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}} & {~=~} &{\rm{C}} \\
\end{array}}$
3. So the general solution is:
$\small{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}~=~\rm{C}}$
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 0 when x = 0
• So substituting x = 0 and y = 0 in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}} & {~=~}&{\rm{C}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\frac{\mathrm{e}^{0}}{4}~+~\frac{e^{0}}{3}} & {~=~}&{\rm{C}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\frac{1}{4}~+~\frac{1}{3}} & {~=~}&{\rm{C}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\rm{C}} & {~=~} &{\frac{7}{12}} \\
\end{array}}$
5. So the particular solution is:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{\mathrm{e}^{-4y}}{4}~+~\frac{e^{3x}}{3}} & {~=~}&{\frac{7}{12}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{3 e^{-4x}~+~4 e^{3x}} & {~=~}&{7} \\
\end{array}}$
Solved example 25.92
Solve the differential
equation
$\small{(x dy - y dx)y\,\sin\left(\frac{y}{x} \right)~=~(y dx + x dy)x\,\cos\left(\frac{y}{x} \right)}$.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{(x dy - y dx)y\,\sin\left(\frac{y}{x} \right)} & {~=~} &{(y dx + x dy)x\,\cos\left(\frac{y}{x} \right)} \\
{~\color{magenta} 2 } &{\Rightarrow} &{xy\,dy\,\sin\left(\frac{y}{x} \right)~-~y^2\,dx\,\sin\left(\frac{y}{x} \right)} & {~=~} &{xy\,dx\,\cos\left(\frac{y}{x} \right)~+~x^2\,dy\,\cos\left(\frac{y}{x} \right)} \\
{~\color{magenta} 3 } &{\Rightarrow} &{xy\,dy\,\sin\left(\frac{y}{x} \right)~-~x^2\,dy\,\cos\left(\frac{y}{x} \right)} & {~=~} &{xy\,dx\,\cos\left(\frac{y}{x} \right)~+~y^2\,dx\,\sin\left(\frac{y}{x} \right)} \\
{~\color{magenta} 4 } &{\Rightarrow} &{\left[xy\,\sin\left(\frac{y}{x} \right)~-~x^2\,\cos\left(\frac{y}{x} \right) \right]dy} & {~=~} &{\left[xy\,\cos\left(\frac{y}{x} \right)~+~y^2\,\sin\left(\frac{y}{x} \right) \right]dx} \\
{~\color{magenta} 5 } &{\Rightarrow} &{\frac{dy}{dx}} & {~=~} &{\frac{xy\,\cos\left(\frac{y}{x} \right)~+~y^2\,\sin\left(\frac{y}{x} \right) }{xy\,\sin\left(\frac{y}{x} \right)~-~x^2\,\cos\left(\frac{y}{x} \right) }} \\
\end{array}}$
2. Let $\small{F(x,y)~=~\frac{xy\,\cos\left(\frac{y}{x} \right)~+~y^2\,\sin\left(\frac{y}{x} \right) }{xy\,\sin\left(\frac{y}{x} \right)~-~x^2\,\cos\left(\frac{y}{x} \right) }}$
• We need to show that, this is a homogeneous function of degree zero.
(i) We can rearrange F(x,y) as shown below:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{F(x,y)} & {~=~} &{\frac{xy\,\cos\left(\frac{y}{x} \right)~+~y^2\,\sin\left(\frac{y}{x} \right) }{xy\,\sin\left(\frac{y}{x} \right)~-~x^2\,\cos\left(\frac{y}{x} \right) }} \\
{~\color{magenta} 2 } &{\Rightarrow} &{F(x,y)} & {~=~} &{\left[\frac{x^2}{x^2} \right]\left[\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) } \right]} \\
{~\color{magenta} 3 } &{} &{} & {~=~} &{\left[1 \right]\left[\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) } \right]} \\
{~\color{magenta} 4 } &{} &{} & {~=~} &{x^0\left[\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) } \right]} \\
{~\color{magenta} 5 } &{} &{} & {~=~} &{x^0\left[g\left(\frac{y}{x} \right) \right]} \\
\end{array}}$
Here, $\small{g\left(\frac{y}{x} \right)~=\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) }}$
(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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{\left(\frac{y}{x} \right)\cos\left(\frac{y}{x} \right)~+~\left(\frac{y}{x} \right)^2\sin\left(\frac{y}{x} \right) }{\left(\frac{y}{x} \right)\sin\left(\frac{y}{x} \right)~-~\cos\left(\frac{y}{x} \right) }} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{v~+~x \frac{dv}{dx}} & {~=~} &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }~-~v} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{(v)\cos\left(v \right)~+~\left(v \right)^2\sin\left(v \right)~-~\left(v \right)^2\sin\left(v \right)~+~(v)\cos(v) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{(v)\cos\left(v \right)~+~(v)\cos(v) }{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{2(v)\cos\left(v \right)}{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right) }} \\
{~\color{magenta} 7 } &{{\Rightarrow}} &{\left[\frac{\left(v \right)\sin\left(v \right)~-~\cos\left(v \right)}{2(v)\cos(v)} \right]dv} & {~=~} &{\left[\frac{1}{x} \right]dx} \\
{~\color{magenta} 8 } &{{\Rightarrow}} &{\left[\frac{\tan(v)}{2}~-~\frac{1}{2(v)} \right]dv} & {~=~} &{\left[\frac{1}{x} \right]dx} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[\frac{\tan(v)}{2}~-~\frac{1}{2(v)} \right]dv}} & {~=~} &{\int{\left[\frac{1}{x} \right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{\frac{\log\left( \left|\sec v \right| \right)}{2}~-~\frac{\log\left( \left|v \right| \right)}{2} ~+~\rm{C}_1} & {~=~}&{\log \left|x \right|~+~\rm{C}_2} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{\log\left( \left|\sec v \right| \right)~-~\log\left( \left|v \right| \right)~+~2{\rm{C}_1}} & {~=~}&{2\log \left|x \right|~+~2\rm{C}_2} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{\log\left( \left|\frac{\sec v}{v x^2} \right| \right)} & {~=~}&{\rm{C}_3} \\
{~\color{magenta} 5 }&{{\Rightarrow}} &{\log\left( \left|\frac{\sec v}{v x^2} \right| \right)} & {~=~}&{\log\left( \left|{\rm{C}_4} \right| \right)} \\
{~\color{magenta} 6 }&{{\Rightarrow}} &{\frac{\sec v}{v x^2}} & {~=~}&{\pm{\rm{C}_4}} \\
\end{array}}$
• The reader may write all steps related to the integration in [(2) magenta color]
5. Replacing 'v', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{\sec v}{v x^2}} & {~=~} &{\rm{C}_4} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{\frac{\sec(y/x)}{(y/x)x^2}} & {~=~} &{\rm{C}_4} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{\frac{\sec(y/x)}{xy}} & {~=~} &{\rm{C}_4} \\
{~\color{magenta} 4 }&{{\Rightarrow}}&{\sec(y/x)} & {~=~} &{{\rm{C}}xy} \\
\end{array}}$
Solved example 25.93
Verify that the function $\small{y ~=~c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)}$, where $\small{c_1,~c_2}$ are arbitrary constants is a solution of the differential
equation
$\small{\frac{dy}{dx}~-~2a\,\frac{dy}{dx}~+~\left(a^2 + b^2 \right)y~=~0}$.
Solution:
1. Find the derivatives from the given solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{c_1\left[e^{ax}(-\sin bx)(b)+(\cos bx)e^{ax}(a) \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~c_2\left[e^{ax}(\cos bx)(b)+(\sin bx)e^{ax}(a) \right]} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{{\frac{dy}{dx}}} & {~=~} &{e^{ax}(\sin bx)\left[c_2\,a~-~c_1\,b \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[c_2\,b~+~c_1\,a \right]} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{\left[c_2\,a~-~c_1\,b \right]\left[e^{ax}(\cos bx)(b)+(\sin bx)e^{ax}(a) \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+\left[c_2\,b~+~c_1\,a \right]\left[e^{ax}(-\sin bx)(b)+(\cos bx)e^{ax}(a) \right]} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{e^{ax}(\sin bx)\Big[\left[c_2\,a~-~c_1\,b \right]a~-~\left[c_2\,b~+~c_1\,a \right]b \Big] } \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~e^{ax}(\cos bx)\Big[\left[c_2\,a~-~c_1\,b \right]b~+~\left[c_2\,b~+~c_1\,a \right]a \Big]} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{{\frac{d^2y}{dx^2}}} & {~=~} &{e^{ax}(\sin bx)\Big[c_2\left(a^2 - b^2 \right)~-~2c_1\,ab \Big] } \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~e^{ax}(\cos bx)\Big[2c_2\,ab~+~c_1\left(a^2 - b^2 \right) \Big] } \\
\end{array}}$
2. Let us write $\small{2a\,\frac{dy}{dx}}$:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{e^{ax}(\sin bx)\left[c_2\,a~-~c_1\,b \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[c_2\,b~+~c_1\,a \right]} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{{2a\,\frac{dy}{dx}}} & {~=~} &{e^{ax}(\sin bx)\left[2c_2\,a^2~-~2c_1\,ab \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[2c_2\,ab~+~2c_1\,a^2 \right]} \\
\end{array}}$
3. Let us write $\small{\left(a^2~+~b^2 \right)y}$:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{c_1\,e^{ax}\,\cos(bx)~+~c_2\,e^{ax}\,\sin(bx)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{\left(a^2~+~b^2 \right)y} & {~=~} &{c_1\,a^2\,e^{ax}\,\cos(bx)~+~c_2\,a^2\,e^{ax}\,\sin(bx)} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+c_1\,b^2\,e^{ax}\,\cos(bx)~+~c_2\,b^2\,e^{ax}\,\sin(bx)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{\left(a^2~+~b^2 \right)y} & {~=~} &{e^{ax}(\sin bx)\left[c_2\,a^2~+~c_2\,b^2 \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[c_1\,a^2~+~c_1\,b^2 \right]} \\
\end{array}}$
4. Let us write all three terms of the differential equation:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{d^2y}{dx^2}} & {~=~} &{e^{ax}(\sin bx)\Big[c_2\left(a^2 - b^2 \right)~-~2c_1\,ab \Big]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~e^{ax}(\cos bx)\Big[2c_2\,ab~+~c_1\left(a^2 - b^2 \right) \Big] } \\
{~\color{magenta} 2 } &{{}} &{-2a\,\frac{dy}{dx}} & {~=~} &{-e^{ax}(\sin bx)\left[2c_2\,a^2~-~2c_1\,ab \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{-e^{ax}(\cos bx)\left[2c_2\,ab~+~2c_1\,a^2 \right]} \\
{~\color{magenta} 3 } &{{}} &{\left(a^2~+~b^2 \right)y} & {~=~} &{e^{ax}(\sin bx)\left[c_2\,a^2~+~c_2\,b^2 \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[c_1\,a^2~+~c_1\,b^2 \right]} \\
\end{array}}$
4. Let us number the similar terms:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{d^2y}{dx^2}} & {~=~} &{e^{ax}(\sin bx)\Big[c_2\left(a^2(I) - b^2(II) \right)~-~2c_1\,ab(III) \Big]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+~e^{ax}(\cos bx)\Big[2c_2\,ab(IV)~+~c_1\left(a^2(V) - b^2(VI) \right) \Big] } \\
{~\color{magenta} 2 } &{{}} &{-2a\,\frac{dy}{dx}} & {~=~} &{-e^{ax}(\sin bx)\left[2c_2\,a^2(I)~-~2c_1\,ab(III) \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{-e^{ax}(\cos bx)\left[2c_2\,ab(IV)~+~2c_1\,a^2(V) \right]} \\
{~\color{magenta} 3 } &{{}} &{\left(a^2~+~b^2 \right)y} & {~=~} &{e^{ax}(\sin bx)\left[c_2\,a^2(I)~+~c_2\,b^2(II) \right]} \\
{~\color{magenta} {} } &{{}} &{{}} & {{}} &{+e^{ax}(\cos bx)\left[c_1\,a^2(V)~+~c_1\,b^2(VI) \right]} \\
\end{array}}$
5. We see that, the terms with the same numbers, cancel each other. So the net sum is zero. That means, L.H.S is equal to R.H.S.
Solved example 25.94
Solve the differential
equation
$\small{(\tan^{-1}y~-~x)dy~=~(1~+~y^2)dx}$.
Solution:
1. The given differential equation is:
$\small{(\tan^{-1}y~-~x)dy~=~(1~+~y^2)dx}$
This can be rearranged as: $\small{\frac{dy}{dx}~=~\frac{1~+~y^2}{\tan^{-1}y~-~x}}$
It is not a linear form. So we rearrange in another way: $\small{\frac{dx}{dy}~=~\frac{\tan^{-1}y~-~x}{1~+~y^2}}$
$\small{\Rightarrow \frac{dx}{dy}~=~\frac{\tan^{-1}y}{1~+~y^2}~-~\frac{x}{1~+~y^2}}$
$\small{\Rightarrow \frac{dx}{dy}~+~\left(\frac{1}{1~+~y^2} \right)x~=~\frac{\tan^{-1}y}{1~+~y^2}}$
• Here P1 = $\small{\frac{1}{1~+~y^2}}$ and Q1 = $\small{\frac{\tan^{-1}y}{1~+~y^2}}$
• P1 and Q1 must satisfy any one of the following three conditions:
(i) Both P1 and Q1 are constants
(ii) Both P1 and Q1 are functions of y
(iii) P1 is a constant and Q1 is a function of y
• Here, both P1 and Q1 are functions of y. So it is a first order linear differential equation.
2. Finding I.F = $\small{e^{\left(\int{\left[{\rm{P}_1} \right]dy} \right)}}$
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{\int{\left[{\rm{P}_1} \right]dy}} & {~=~} &{\int{\left[\frac{1}{1~+~y^2} \right]dy}~=~\tan^{-1}y} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{e^{\left(\int{\left[{\rm{P}_1} \right]dx} \right)}} & {~=~} &{e^{\left(\tan^{-1}y \right)}} \\
\end{array}}$
3. Write the equation with independent x:
$\small{x\left[{\rm{I.F}} \right]~=~\int{\left[{\rm{Q_1(I.F)}} \right]dy}}$
We get: $\small{x\left[e^{\left(\tan^{-1}y \right)} \right]~=~\int{\left[\frac{\tan^{-1}y}{1~+~y^2}~\times~e^{\left(\tan^{-1}y \right)} \right]dy}}$
4. Do the integration in the R.H.S, and add the constant of integration.
We get: $\small{e^{\left(\tan^{-1}y \right)}\left[\tan^{-1}y~-~1 \right]~+~\rm{C}}$
5. Multiply the above result in (4), by the reciprocal of the I.F
$\small{\begin{array}{ll}{~\color{magenta} 1 } &{{}} &{x} & {~=~} &{\left[e^{\left(\tan^{-1}y \right)}\left[\tan^{-1}y~-~1 \right]~+~\rm{C} \right]\frac{1}{e^{\left(\tan^{-1}y \right)}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x} & {~=~} &{\tan^{-1}y~-~1~+~{\rm{C}}\,e^{\left(-\tan^{-1}y \right)}} \\
\end{array}}$
The link below gives a few more miscellaneous examples:
Miscellaneous Exercise
In the next chapter, we will see Vector Algebra.
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