# How do you find the particular solution to (du)/(dv)=uvsinv^2 that satisfies u(0)=1?

Jan 11, 2017

$u \left(v\right) = {e}^{{\sin}^{2} \left({v}^{2} / 2\right)}$

#### Explanation:

We can find the general solution by separating the variables:

$\frac{\mathrm{du}}{\mathrm{dv}} = u v \sin \left({v}^{2}\right)$

$\frac{\mathrm{du}}{u} = v \sin \left({v}^{2}\right) \mathrm{dv}$

$\int \frac{\mathrm{du}}{u} = \int v \sin \left({v}^{2}\right) \mathrm{dv}$

$\ln u = \frac{1}{2} \int \sin \left({v}^{2}\right) d \left({v}^{2}\right) = - \frac{1}{2} \cos \left({v}^{2}\right) + C$

To determine $C$ we note that for $v = 0$ we have:

$u \left(0\right) = 1 \implies \ln \left(u \left(0\right)\right) = 0$

so

$- \frac{1}{2} \cos \left(0\right) + C = 0$

$- \frac{1}{2} + C = 0$

$C = \frac{1}{2}$

The particular solution we look for is then:

$\ln u = - \frac{1}{2} \cos \left({v}^{2}\right) + \frac{1}{2} = \frac{1}{2} \left(1 - \cos \left({v}^{2}\right)\right) = {\sin}^{2} \left({v}^{2} / 2\right)$

$u \left(v\right) = {e}^{{\sin}^{2} \left({v}^{2} / 2\right)}$