Find the particular solution of the differential equation x ( y x ) dy dx =y ( y x )+x, given that when x=1,y= 4.

x ( y x ) dy dx =y ( y x )+x dy dx = y ( y x )+x x ( y x ) This is a homogeneous differential equation. puttuing y = vx and dy dx =v + x dv dx , we get v + x dv dx = vx+ + x x v + x dv dx = v+ + 1 x dv dx = v +1-v v x dv dx =1 dv=1 x dx Integrating both sides, we get dv= 1 x dx = |x|+C Putting v= y x , we get y x = |x|+C (1) At x=1,y= 4 (Given) Putting x=1 and y= 4 in (1), we get C=1 2 Putting C=1 2 in (1), we get y x = |x|+1 2 Hence, y x = |x|+1 2 is the required solution.

Find the particular solution of the differential equation x ( y x…

Download App

Question

Find the particular solution of the differential equation $\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x},$ given that when $\text{x}=1,\text{y}=\frac{\pi}4$.

✓

Answer

$\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)\frac{\text{dy}}{\text{dx}}=\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{y}\cos\Big(\frac{\text{y}}{\text{x}}\Big)+\text{x}}{\text{x}\cos\Big(\frac{\text{y}}{\text{x}}\Big)}$
This is a homogeneous differential equation.
puttuing y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}},$ we get
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}+\cos\text{v + x}}{\text{x}\cos\text{v}}$
$\Rightarrow\ \text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}+\cos\text{v + 1}}{\cos\text{v}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{\text{v}\cos\text{v}+1-\text{v}\cos\text{v}}{\cos{\text{v}}}$
$\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1}{\cos\text{v}}$
$\Rightarrow\ \cos\text{v dv}=\frac{1}{\text{x}}\text{dx}$
Integrating both sides, we get
$\int\cos\text{v dv}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow\ \sin\text{v}=\log|\text{x}|+\text{C}$
Putting $\text{v}=\frac{\text{y}}{\text{x}},$ we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\text{C}\ \dots(1)$
At $\text{x}=1,\text{y}=\frac{\pi}4$ (Given)
Putting $\text{x}=1$ and $\text{y}=\frac{\pi}4$ in (1), we get
$\text{C}=\frac{1}{\sqrt2}$
Putting $\text{C}=\frac{1}{\sqrt2}$ in (1), we get
$\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$
Hence, $\sin\frac{\text{y}}{\text{x}}=\log|\text{x}|+\frac{1}{\sqrt2}$ is the required solution.