In the previous section,
we saw the method to solve homogeneous differential equations. We also saw some solved examples. In this
section, we will see a few more solved examples.
Solved example 25.52
Show that the differential equation $\boldsymbol{x
\cos\left(\frac{y}{x} \right) \,\frac{dy}{dx}~=~y \cos \left(\frac{y}{x}
\right)~+~x}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\frac{dy}{dx}~=~\frac{y \cos \left(\frac{y}{x} \right)~+~x}{x \cos\left(\frac{y}{x} \right)}}$
2. Let $\small{F(x,y)~=~\frac{y \cos \left(\frac{y}{x} \right)~+~x}{x \cos\left(\frac{y}{x} \right)}}$
• We need to show that, this is a homogeneous function of degree zero.
3. Dividing both numerator and denominator by x, we get:
$\small{F(x,y)~=~\frac{\left(\frac{y}{x}
\right) \cos \left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x}
\right)}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$
• Here,
$\boldsymbol{g\left(\frac{y}{x} \right) ~=~\frac{\left(\frac{y}{x}
\right) \cos \left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x}
\right)}}$
4. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g\left(\frac{y}{x} \right) \right]}$
♦ Where n is a natural number.
• So it is a homogeneous function.
• For this function, n = 0. So degree is zero.
• So 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:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}}
& {~=~} &{\frac{\left(\frac{y}{x} \right) \cos
\left(\frac{y}{x} \right)~+~1}{\cos\left(\frac{y}{x} \right)}} \\
{~\color{magenta}
2 } &{{\Rightarrow}} &{v~+~x \frac{dv}{dx}} &
{~=~} &{\frac{\left(v \right) \cos \left(v
\right)~+~1}{\cos\left(v \right)}} \\
{~\color{magenta} 3
} &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~}
&{\frac{v \cos \left(v \right)~+~1}{\cos\left(v \right)}~-~v} \\
{~\color{magenta}
4 } &{{\Rightarrow}} &{x \frac{dv}{dx}} &
{~=~} &{\frac{v \cos \left(v \right)~+~1~-~v \cos\left(v
\right)}{\cos\left(v \right)}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{1}{\cos v}} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\left[\cos v \right]dv} & {~=~} &{\frac{dx}{x}} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}}&{\int{\left[\cos v
\right]dv}} & {~=~} &{\int{\left[\frac{1}{x}
\right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}}
&{\sin v~+~\rm{C}_1} & {~=~}&{\log \left|x
\right|~+~\rm{C}_2} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{\sin v} & {~=~}&{\log \left|x \right|~+~\rm{C}_3} \\
\end{array}}$
5. Replacing 'v', we will get the general solution:
$\small{\begin{array}{ll}
{~\color{magenta} 1 } &{{}}&{\sin v} & {~=~}
&{\log \left|x \right|~+~\rm{C}_3} \\
{~\color{magenta}
2 }&{{\Rightarrow}}&{\sin\left(\frac{y}{x} \right)} &
{~=~} &{\log \left|x \right|~+~\log \left|\rm{C} \right|} \\
{~\color{magenta}
3 }&{{\Rightarrow}}&{\sin\left(\frac{y}{x} \right)} &
{~=~} &{\log \left|\rm{C}\,x \right|} \\
\end{array}}$
Solved example 25.53
Show that the differential equation $\boldsymbol{x \, \frac{dy}{dx}~-~y~+~x \sin \left(\frac{y}{x} \right)~=~0}$ is homogeneous and solve it.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{x \, \frac{dy}{dx}~-~y~+~x \sin \left(\frac{y}{x} \right)} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{x \, \frac{dy}{dx}} & {~=~} &{y~-~x \sin \left(\frac{y}{x} \right)} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{ \frac{dy}{dx}} & {~=~} &{\frac{y~-~x \sin \left(\frac{y}{x} \right)}{x}} \\
\end{array}}$
2. Let $\small{F(x,y)~=~\frac{y~-~x \sin \left(\frac{y}{x} \right)}{x}}$
• We need to show that, this is a homogeneous function of degree zero.
3. Dividing both numerator and denominator by x, we get:
$\small{F(x,y)~=~\frac{\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)}{1}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$
• Here, $\boldsymbol{g\left(\frac{y}{x} \right) ~=~\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)}$
4. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g\left(\frac{y}{x} \right) \right]}$
♦ Where n is a natural number.
• So it is a homogeneous function.
• For this function, n = 0. So degree is zero.
• So 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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\left(\frac{y}{x} \right)~-~ \sin \left(\frac{y}{x} \right)} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{v~+~x \frac{dv}{dx}} & {~=~} &{v~-~\sin v} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{-\sin v} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\left[\frac{1}{\sin v} \right]dv} & {~=~} &{-\frac{dx}{x}} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[\frac{1}{\sin v} \right]dv}} & {~=~} &{(-1)\int{\left[\frac{1}{x} \right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{\log \left|\csc v~-~\cot v \right|~+~\rm{C}_1} & {~=~}&{(-1)\log \left|x \right|~+~\rm{C}_2} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{\log \left|\csc v~-~\cot v \right|~+~\log \left|x \right|} & {~=~}&{\rm{C}_2~-~\rm{C}_1} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{\log \left|\csc v~-~\cot v \right|~+~\log \left|x \right|} & {~=~}&{\rm{C}_3} \\
{~\color{magenta} 5 }&{{\Rightarrow}} &{\log \left|(\csc v~-~\cot v)x \right|} & {~=~}&{\rm{C}_3} \\
{~\color{magenta} 6 }&{{\Rightarrow}} &{(\csc v~-~\cot v)x} & {~=~}&{e^{\rm{C}_3}~=~\rm{C}} \\
{~\color{magenta} 7 }&{{\Rightarrow}} &{\frac{x(1~-~\cos v)}{\sin v}} & {~=~}&{e^{\rm{C}_3}~=~\rm{C}} \\
\end{array}}$
5. Replacing 'v', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\frac{x(1~-~\cos v)}{\sin v}} & {~=~} &{\rm{C}} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{\frac{x\left[1~-~\cos \left(\frac{y}{x} \right) \right]}{\sin \left(\frac{y}{x} \right)}} & {~=~} &{\rm{C}} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{x\left[1~-~\cos \left(\frac{y}{x} \right) \right]} & {~=~} &{\rm{C}\sin\left(\frac{y}{x} \right)} \\
\end{array}}$
Solved example 25.54
Show that the differential equation $\boldsymbol{2y e^{x/y} dx~+~\left(y\,-\,2x e^{x/y} \right)dy~=~0}$ is
homogeneous and find it's particular solution, given that, x = 0 when y = 1.
Solution:
Part I: Showing that the given differential equation is homogeneous
1. We can rearrange the given differential equation as:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{2y e^{x/y} dx~+~\left(y\,-\,2x e^{x/y} \right)dy} & {~=~} &{0} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{2y e^{x/y} dx} & {~=~} &{(-1)\left(y\,-\,2x e^{x/y} \right)dy} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{2y e^{x/y} dx} & {~=~} &{\left(2x e^{x/y}\,-\,y \right)dy} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{\frac{dx}{dy}} & {~=~} &{\frac{2x e^{x/y}\,-\,y }{2y e^{x/y}}} \\
\end{array}}$
• Here we write $\small{\frac{dx}{dy}}$ instead of $\small{\frac{dy}{dx}}$ because, the given differential equation involves $\small{x/y}$.
2. Let $\small{F(x,y)~=~\frac{2x e^{x/y}\,-\,y }{2y e^{x/y}}}$
• We need to show that, this is a homogeneous function of degree zero.
3. Dividing both numerator and denominator by y, we get:
$\small{F(x,y)~=~\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}~=~y^0\left[h\left(\frac{x}{y} \right) \right]}$
• Here, $\boldsymbol{h\left(\frac{x}{y} \right) ~=~\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}}$
4. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~y^n\left[h\left(\frac{x}{y} \right) \right]}$
♦ Where n is a natural number.
• So it is a homogeneous function.
• For this function, n = 0. So degree is zero.
• So the given differential equation is homogeneous.
Part II: Solving the differential equation
1. We make the substitution: $\small{\frac{x}{y}~=~u}$
Which is same as: $\small{x~=~uy}$
2. Differentiating the above equation with respect to y, we get:
$\small{\frac{dx}{dy}~=~u~+~y \frac{du}{dy}}$
3. Substituting (1) and (2) in the given differential equation, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dx}{dy}} & {~=~} &{\frac{2(x/y) e^{x/y}\,-\,1 }{2 e^{x/y}}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{u~+~y \frac{du}{dy}} & {~=~} &{\frac{2(u) e^{u}\,-\,1 }{2 e^{u}}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{u~+~y \frac{du}{dy}} & {~=~} &{u~-~\frac{1 }{2 e^{u}}} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{y \frac{du}{dy}} & {~=~} &{-\frac{1 }{2 e^{u}}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{2 e^u\,du} & {~=~} &{-\frac{1 }{y}dy} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[2 e^u \right]du}} & {~=~} &{(-1)\int{\left[\frac{1}{y} \right]dy}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{2 e^u~+~\rm{C}_1} & {~=~}&{(-1)\log \left|y \right|~+~\rm{C}_2} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{2 e^u~+~\log \left|y \right|} & {~=~}&{\rm{C}_2~-~\rm{C}_1} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{2 e^u~+~\log \left|y \right|} & {~=~}&{\rm{C}} \\
\end{array}}$
5. Replacing 'u', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{2 e^u~+~\log \left|y \right|} & {~=~} &{\rm{C}} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{2 e^{(x/y)}~+~\log \left|y \right|} & {~=~} &{\rm{C}} \\
\end{array}}$
Part III: To find the particular solution
1. Given that, x = 0 when y = 1.
2. Substituting in the general solution, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{2 e^{(x/y)}~+~\log \left|y \right|} & {~=~} &{\rm{C}} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{2 e^{(0/1)}~+~\log \left(1 \right)} & {~=~} &{\rm{C}} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{2 e^{(0)}~+~0} & {~=~} &{\rm{C}} \\
{~\color{magenta} 4 }&{{\Rightarrow}}&{2 (1)~+~0} & {~=~} &{\rm{C}} \\
{~\color{magenta} 5 }&{{\Rightarrow}}&{2} & {~=~} &{\rm{C}} \\
\end{array}}$
3. So based on the result from part II, we get the particular solution as:
$\boldsymbol{2 e^{(x/y)}~+~\log \left|y \right|~=~2}$
Solved example 25.55
Show that the family of curves for which the slope of the tangent at any point (x,y) on it is $\boldsymbol{\frac{x^2 + y^2}{2xy}}$, is given by $\boldsymbol{x^2 ~-~y^2~=~cx}$.
Solution:
Part I: Finding the differential equation
1. We know that, the slope is same as the derivative $\small{\frac{dy}{dx}}$.
2. So based on the given information, we can write:
$\small{\frac{dy}{dx}~=~\frac{x^2 + y^2}{2xy}}$
• This is the required differential equation.
Part II: Showing that the differential equation is homogeneous
1. We have the differential equation:
$\small{\frac{dy}{dx}~=~\frac{x^2 + y^2}{2xy}}$
2. Let $\small{F(x,y)~=~\frac{x^2 + y^2}{2xy}}$
• We need to show that, this is a homogeneous function of degree zero.
3. Dividing both numerator and denominator by $\small{x^2}$, we get:
$\small{F(x,y)~=~\frac{1 + \frac{y^2}{x^2}}{2\left(\frac{y}{x} \right)}~=~x^0\left[g\left(\frac{y}{x} \right) \right]}$
• Here, $\boldsymbol{g\left(\frac{y}{x} \right) ~=~\frac{1 + \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)}}$
4. We see that:
For the function $\small{F(x,y)}$, it is possible to write:
$\small{F(x,y)~=~x^n\left[g\left(\frac{y}{x} \right) \right]}$
♦ Where n is a natural number.
• So it is a homogeneous function.
• For this function, n = 0. So degree is zero.
• So the given differential equation is homogeneous.
Part III: 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:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{1 + \left(\frac{y}{x} \right)^2}{2\left(\frac{y}{x} \right)}} \\
{~\color{magenta} 2 } &{{\Rightarrow}} &{v~+~x \frac{dv}{dx}} & {~=~} &{\frac{1 + \left(v \right)^2}{2\left(v \right)}~=~\frac{1}{2v}~+~\frac{v}{2}} \\
{~\color{magenta} 3 } &{{\Rightarrow}} &{x \frac{dv}{dx}} & {~=~} &{\frac{1}{2v}~-~\frac{v}{2}~=~\frac{1}{2}\left[\frac{1}{v}~-~v \right]} \\
{~\color{magenta} 4 } &{{\Rightarrow}} &{ \frac{dv}{\frac{1}{v}~-~v }} & {~=~} &{\frac{dx}{2x}} \\
{~\color{magenta} 5 } &{{\Rightarrow}} &{\left[ \frac{1}{\frac{1}{v}~-~v } \right]dv} & {~=~} &{\left[ \frac{1}{2x } \right]dx} \\
{~\color{magenta} 6 } &{{\Rightarrow}} &{\left[\frac{v}{1~-~v^2 } \right]dv} & {~=~} &{\left[ \frac{1}{2x } \right]dx} \\
\end{array}}$
4. So we have separated the variables. Integrating both sides, we get:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{\int{\left[\frac{v}{1~-~v^2 } \right]dv}} & {~=~} &{\int{\left[\frac{1}{2x} \right]dx}} \\
{~\color{magenta} 2 }&{{\Rightarrow}} &{(-1)\left(\frac{1}{2} \right)\log \left|(1 - v^2) \right|~+~\rm{C}_1} & {~=~}&{\left(\frac{1}{2} \right)\log \left|x \right|~+~\rm{C}_2} \\
{~\color{magenta} 3 }&{{\Rightarrow}} &{(-1)\log \left|(1 - v^2) \right|~+~2\rm{C}_1} & {~=~}&{\log \left|x \right|~+~2\rm{C}_2} \\
{~\color{magenta} 4 }&{{\Rightarrow}} &{\log \left|x \right|~+~\log \left|(1 - v^2) \right|} & {~=~}&{2\rm{C}_1~-~2\rm{C}_2} \\
{~\color{magenta} 5 }&{{\Rightarrow}} &{\log \left|x(1 - v^2) \right|} & {~=~}&{\rm{C}_3} \\
{~\color{magenta} 6 }&{{\Rightarrow}} &{\log \left|x(1 - v^2) \right|} & {~=~}&{\log \left|\rm{C}_4 \right|} \\
{~\color{magenta} 7 }&{{\Rightarrow}} &{x(1 - v^2)} & {~=~}&{\pm\rm{C}_4} \\
\end{array}}$
5. Replacing 'v', we will get the general solution:
$\small{\begin{array}{ll} {~\color{magenta} 1 } &{{}}&{x(1 - v^2)} & {~=~} &{\pm \rm{C}_4} \\
{~\color{magenta} 2 }&{{\Rightarrow}}&{x\left(1 - \frac{y^2}{x^2}\right) } & {~=~} &{\pm \rm{C}_4} \\
{~\color{magenta} 3 }&{{\Rightarrow}}&{\frac{x^2~-~y^2}{x}} & {~=~} &{\pm \rm{C}_4} \\
{~\color{magenta} 4 }&{{\Rightarrow}}&{x^2~-~y^2} & {~=~} &{{\rm{C}}\,x} \\
\end{array}}$
The link below gives a few more solved examples:
Exercise 9.5
In the next section, we will see linear differential equations.
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