Solve the following differential equation: x ( y x ).(y dx + x dy)=y ( y x ).(x dy-y dx)

We have, x ( y x ).(y dx + x dy)=y ( y x ).(x dy-y dx) xy ( y x )dx+x^2 ( y x )dy=xy ( y x )dy-y^2 ( y x )dx [xy ( y x )+y^2 ( y x ) ]dx= [xy ( y x )-x^2 ( y x ) ]dy dy dx = xy ( y x )+y^2 ( y x ) xy ( y x )-x^2 ( y x ) This is a homogeneous differential equation. Putting x = vy and dy dx =v + x dv dx , we get v + x dv dx = vx^2 +v^2x^2 vx^2 -x^2 v + x dv dx = v +v^2 v - x dv dx = v +v^2 v - -v x dv dx = v +v^2 -v^2 +v v - x dv dx = 2v v - v - 2v dv=1 x dx Integrating both sides, we get v - 2v dv= 1 x dx v - v dv=2 1 x dx v v dv- v dv=2 1 x dx dv- 1 v dv=2 1 x dx | |- |v|=2 |x|+ | v |= |Cx^2 | v v =Cx^2 Putting v= y x , we get ( y x )= y x ^2 ( y x )=Cxy Hence, ( y x )=Cxy is the required solution.

Solve the following differential equation: x ( y x ).(y dx + x dy…

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Question

Solve the following differential equation:
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$

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Answer

We have,
$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big).(\text{y dx + x dy})=\text{y}\sin\Big(\frac{\text{y}}{\text{x}}\Big).(\text{x dy}-\text{y dx})$
$\Rightarrow\ \text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}+\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}=\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dy}-\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\text{dx}$
$\Rightarrow\ \Big[\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dx}=\Big[\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)\Big]\text{dy}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{xy}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{y}^2\sin\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{xy}\sin\Big(\frac{\text{y}}{\text{x}}\Big)-\text{x}^2\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
Putting x = vy and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}^2\cos\text{v}+\text{v}^2\text{x}^2\sin\text{v}}{\text{vx}^2\sin\text{v}-\text{x}^2\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}-\text{v}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+\text{v}^2\sin\text{v}-\text{v}^2\sin\text{v}+\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{2\text{v}\cos\text{v}}{\text{v}\sin\text{v}-\cos\text{v}}$
$\Rightarrow\ \frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\frac{\text{v}\sin\text{v}-\cos\text{v}}{2\text{v}\cos\text{v}}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}-\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\frac{\text{v}\sin\text{v}}{\text{v}\cos\text{v}}\text{dv}-\int\frac{\cos\text{v}}{\text{v}\cos\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \int\tan\text{v dv}-\int\frac{1}{\text{v}}\text{dv}=2\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \log|\sec\text{v}|-\log|\text{v}|=2\log|\text{x}|+\log\text{C}$
$\Rightarrow\ \log\Big|\frac{\sec\text{v}}{\text{v}}\Big|=\log\big|\text{Cx}^2\big|$
$\Rightarrow\ \frac{\sec{\text{v}}}{\text{v}}=\text{Cx}^2$
Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get
$\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\frac{\text{y}}{\text{x}}\times\text{C}\times\text{x}^2$
$\Rightarrow\ \sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$
Hence, $\sec\Big(\frac{\text{y}}{\text{x}}\Big)=\text{Cxy}$ is the required solution.

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