1. ## Differential equations substitution

Solve:

$xyy'+y^2=sinx$ by letting $y^2=v$

Would you use this and then substitute it into the differential equation?

$y'=\dfrac{1}{2\sqrt{v}}\dfrac{dv}{dx}$

2. $y^2 = v$

$y = \sqrt{v}$

$\frac{d}{dx}(y) = \frac{d}{dx}(\sqrt{v})$

$\frac{dy}{dx} = \frac{d}{dv}(\sqrt{v})\,\frac{dv}{dx}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{v}}\,\frac{dv}{dx}$.

So yes, I would substitute this into the DE...

$x\sqrt{v}\left(\frac{1}{2\sqrt{v}}\,\frac{dv}{dx}\ right) + v = \sin{x}$

$\frac{1}{2}x\,\frac{dv}{dx} + v = \sin{x}$

$\frac{dv}{dx} + \frac{2}{x}v = \frac{2}{x}\sin{x}$

Now using the integrating factor

$e^{\int{\frac{2}{x}\,dv}} = e^{2\ln{x}} = e^{\ln{(x^2)}} = x^2$ and multiplying through gives

$x^2\,\frac{dv}{dx} + 2xv = 2x\sin{x}$

$\frac{d}{dx}(x^2v) = 2x\sin{x}$

$x^2v = 2\int{x\sin{x}\,dx}$

$x^2v = 2\left[-x\cos{x} - \int{\cos{x}\,dx}\right]$

$x^2v = 2\left[-x\cos{x} - \sin{x}\right] + C$

$x^2v = -2x\cos{x} - 2\sin{x} + C$

$v = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

$y^2 = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

$y = \pm\sqrt{-\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}}$.

3. Originally Posted by Prove It
$y^2 = v$

$y = \sqrt{v}$

$\frac{d}{dx}(y) = \frac{d}{dx}(\sqrt{v})$

$\frac{dy}{dx} = \frac{d}{dv}(\sqrt{v})\,\frac{dv}{dx}$

$\frac{dy}{dx} = \frac{1}{2\sqrt{v}}\,\frac{dv}{dx}$.

So yes, I would substitute this into the DE...

$x\sqrt{v}\left(\frac{1}{2\sqrt{v}}\,\frac{dv}{dx}\ right) + v = \sin{x}$

$\frac{1}{2}x\,\frac{dv}{dx} + v = \sin{x}$

$\frac{dv}{dx} + \frac{2}{x}v = \frac{2}{x}\sin{x}$

Now using the integrating factor

$e^{\int{\frac{2}{x}\,dv}} = e^{2\ln{x}} = e^{\ln{(x^2)}} = x^2$ and multiplying through gives

$x^2\,\frac{dv}{dx} + 2xv = 2x\sin{x}$

$\frac{d}{dx}(x^2v) = 2x\sin{x}$

$x^2v = 2\int{x\sin{x}\,dx}$

$x^2v = 2\left[-x\cos{x} - \int{\cos{x}\,dx}\right]$

$x^2v = 2\left[-x\cos{x} - \sin{x}\right] + C$

$x^2v = -2x\cos{x} - 2\sin{x} + C$

$v = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

$y^2 = -\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}$

$y = \pm\sqrt{-\frac{2}{x}\cos{x} - \frac{2}{x^2}\sin{x} + \frac{C}{x^2}}$.
Thanks, that makes a lot of sense now