Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
$\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
✓
Answer
Here, $\text{x}\frac{\text{dy}}{\text{dx}}=\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$
It is a homogeneous equation
Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\cos^2\big(\frac{\text{vx}}{\text{x}}\big)}{\text{x}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\cos^2\text{v}$
$\frac{\text{dv}}{\cos^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\int\sec^2\text{vdv}=-\int\frac{\text{dx}}{\text{x}}$
$\tan\text{v}=-\log|\text{x}|+\log|\text{C}|$
$\tan\frac{\text{y}}{\text{x}}=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}-\text{x}\cos^2\Big(\frac{\text{y}}{\text{x}}\Big)}{\text{x}}$
It is a homogeneous equation
Put y = vx
and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{vx}-\text{x}\cos^2\big(\frac{\text{vx}}{\text{x}}\big)}{\text{x}}$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\text{v}-\cos^2\text{v}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=-\cos^2\text{v}$
$\frac{\text{dv}}{\cos^2\text{v}}=-\frac{\text{dx}}{\text{x}}$
$\int\sec^2\text{vdv}=-\int\frac{\text{dx}}{\text{x}}$
$\tan\text{v}=-\log|\text{x}|+\log|\text{C}|$
$\tan\frac{\text{y}}{\text{x}}=\log\Big|\frac{\text{C}}{\text{x}}\Big|$
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