Question

In Problems 45-50 use a technique of integration or a substitution to find an explicit solution of the given differential equation or initial-value problem.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{\mathbf{sin}\sqrt{x}}{\sqrt{y}}$

Short answer

The explicit solution is $y={(-3\sqrt{x}\cos \sqrt{x}+3\sin \sqrt{x}+C)}^{\frac{2}{3}}$.

Step-by-step solution

Definition

A first-order differential equation of the form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{h}\mathbf{\left(}\mathbf{y}\mathbf{\right)}$is said to be separable or to have separable variables.

Separate variable

Consider the differential equation$\frac{dy}{dx}=\frac{\sin \sqrt{x}}{\sqrt{y}}$

Separate variables by cross multiplying.

$\sqrt{y}dy=\sin \sqrt{x}dx$

Now we can integrate the left hand side in terms of $y$and the right hand side in terms of $x$.

$\int \sqrt{y}dy=\int \sin \sqrt{x}dx$

Evaluate integral

Evaluate $\int \sin \sqrt{x}dx$by substituting

$\begin{array}{c}u=\sqrt{x}\\ du=\frac{1}{2\sqrt{x}}\\ du=\frac{1}{2u}dx\end{array}$

$dx=2udu$

Integral

So, we have$\int \sin \sqrt{x}dx=\int 2u\sin udu$

role="math" localid="1667829112287" $\begin{array}{c}\int \sqrt{\mathrm{y}}\mathrm{dy}=\int 2\mathrm{usinudu}\\ \frac{2}{3}{\mathrm{y}}^{\frac{3}{2}}=-2\mathrm{ucosu}+2\mathrm{sinu}+\mathrm{C}\\ {\mathrm{y}}^{\frac{3}{2}}=-3\sqrt{\mathrm{x}}\cos \sqrt{\mathrm{x}}+3\sin \sqrt{\mathrm{x}}+\mathrm{C}\\ y={\left(-3\sqrt{x}Cos\sqrt{x}+3Sin\sqrt{x}+C\right)}^{\frac{2}{3}}\\ \end{array}$