Question

How do you find the particular solution to $\dfrac{{du}}{{dv}} = uv\sin {v^2}$ that satisfies $u\left( 0 \right) = 1$ ?

Answer 1

Here, we are given a differential equation and we are asked to calculate the particular solution to $\dfrac{{du}}{{dv}} = uv\sin {v^2}$ that satisfies $u\left( 0 \right) = 1$
Formula to be used:
a) The power rule of differentiation is $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$
b) $\int {\dfrac{1}{x}} dx = \ln \left| x \right| + C$
Here $C$ is the constant of integration.
c) $\int {\sin xdx = - \cos x + C}$
Here $C$ is the constant of integration.

Complete step by step answer:
The given equation is $\dfrac{{du}}{{dv}} = uv\sin {v^2}$ and we need to find the particular solution of the equation.
Here we need to use the variable separable method. This method involves the following steps.
The first step is to move all the $y$ terms and $dy$ to one side of the equation and all the $x$ terms and $dx$ to another side of the equation. Then we need to integrate them and simplify the equation.
Now, consider $\dfrac{{du}}{{dv}} = uv\sin {v^2}$
$\Rightarrow \dfrac{{du}}{u} = v\sin {v^2}dv$ (Using variable separable method)
Now, apply integral on both sides.
$\int {\dfrac{{du}}{u}} = \int {v\sin {v^2}dv}$ …. $\left( 1 \right)$
Let us consider $t = {v^2}$
Then, $dt = 2vdv$
$\Rightarrow \dfrac{1}{2}dt = vdv$
We shall substitute $\dfrac{1}{2}dt = vdv$ in the equation $\left( 1 \right)$
Thus, we got $\int {\dfrac{{du}}{u}} = \int {\dfrac{1}{2}\sin {t}dt}$
$\Rightarrow \int {\dfrac{{du}}{u}} = \dfrac{1}{2}\int {\sin {t}dt}$
$\Rightarrow \int {\dfrac{{du}}{u}} = - \dfrac{1}{2}\cos {t}$ (Here we applied the formula $\int {\sin xdx = - \cos x + C}$ )
Now put ${v^2} = t$
$\Rightarrow \ln \left| u \right| = - \dfrac{1}{2}\cos {v^2} + C$ ……… $\left( 2 \right)$ (Here we applied the formula $\int {\dfrac{1}{x}} dx = \ln \left| x \right| + C$ )
Here $C$ is the constant of integration.
We need to find the value of $C$
It is given that $u\left( 0 \right) = 1$ that implies $v = 0$ and $u = 1$
Thus, substitute $v = 0$ and $u = 1$ in the equation $\left( 2 \right)$
$\Rightarrow \ln \left| 1 \right| = - \dfrac{1}{2}\cos {\left( 0 \right)^2} + C$
$\Rightarrow 0 = - \dfrac{1}{2} \times 1 + C$ (Here $\ln 1 = 0$ and $\cos 0 = 1$ )
$\Rightarrow C = \dfrac{1}{2}$
Now, we shall apply $C = \dfrac{1}{2}$ in the equation $\left( 2 \right)$
$\ln \left| u \right| = - \dfrac{1}{2}\cos {v^2} + \dfrac{1}{2}$
$\Rightarrow \ln \left| u \right| = \dfrac{{1 - \cos {v^2}}}{2}$
$\Rightarrow {e^{\ln \left| u \right|}} = {e^{\dfrac{{1 - \cos {v^2}}}{2}}}$ (Here we have raised to exponents on both sides)
$\Rightarrow \left| u \right| = {e^{\dfrac{{1 - \cos {v^2}}}{2}}}$
Since ${e^x} > 0,\forall x \in \mathbb{R}$ , the particular solution is
$u = {e^{\dfrac{{1 - \cos {v^2}}}{2}}}$

Note:
The differential can be solved using various methods like integrating factor methods but the key point is to remember where we use which method. Students makes mistake in choosing the method and that makes problem more complicated.