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.

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathbf{xtan}}^{-1}\mathbf{x}}{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{3}$

Short answer

The explicit solution is $y=\sqrt{x^2{\tan }^{-1}x-x+{\tan }^{-1}x+9}$.

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

The differential equation is$\frac{dy}{dx}\mathbf{=}\frac{x{\tan }^{-1}x}{y},y\left(0\right)=3$

Rewrite the differential equation as$ydy=x{\tan }^{-1}xdx$

Integrate on both sides.

$\int \mathrm{ydy}=\int \mathrm{x}{\tan }^{-1}\mathrm{xdx}$

Integrate

In order to solve this integral we use integration by parts$\int wdv=wv-\int vdx$

Let role="math" localid="1667832508183" $v=\mathrm{t}a{n}^{-1}x$and $dv=xdx$.

Then$\mathrm{dw}=\frac{1}{{\mathrm{x}}^2+1}\mathrm{dx},\mathrm{v}=\frac{1}{2}{\mathrm{x}}^2$

The integral becomes,

$\begin{array}{c}\int {\tan }^{-1}\mathrm{xdx}=\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\int \frac{1}{2}{\mathrm{x}}^2\frac{1}{{\mathrm{x}}^2+1}\mathrm{dx}\\ =\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int \frac{{\mathrm{x}}^2}{{\mathrm{x}}^2+1}\mathrm{dx}\mathrm{..........}\left(1\right)\end{array}$

Evaluating integral

The integration becomes,

$\begin{array}{c}\int {\mathrm{xtan}}^{-1}\mathrm{xdx}=\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\frac{1}{2}\int \left(1-\frac{1}{{\mathrm{x}}^2+1}\right)\mathrm{dx}\\ =\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\frac{1}{2}(\mathrm{x}-{\tan }^{-1}\mathrm{x})+\mathrm{c}\\ =\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\frac{1}{2}\mathrm{x}+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\mathrm{c}\end{array}$

From equation (1),$\frac{y^2}{2}=\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}x+\frac{1}{2}{\tan }^{-1}x+c$

Apply initial condition:

Use the initial condition $y\left(0\right)=3$in$\frac{y^2}{2}=\frac{1}{2}{x}^2{\tan }^{-1}x-\frac{1}{2}x+\frac{1}{2}{\tan }^{-1}x+c$

$\frac{3^2}{2}=\frac{1}{2}\left(0^2\right){\tan }^{-1}0-\frac{1}{2}\left(0\right)+\frac{1}{2}{\tan }^{-1}0+\mathrm{c}$

$\frac{9}{2}=c$

So, the solution of differential equation is,

$\begin{array}{c}\frac{{\mathrm{y}}^2}{2}=\frac{1}{2}{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\frac{1}{2}\mathrm{x}+\frac{1}{2}{\tan }^{-1}\mathrm{x}+\frac{9}{2}{\mathrm{y}}^2\\ ={\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\mathrm{x}+{\tan }^{-1}\mathrm{x}+9\mathrm{y}\\ =\sqrt{{\mathrm{x}}^2{\tan }^{-1}\mathrm{x}-\mathrm{x}+{\tan }^{-1}\mathrm{x}+9}\end{array}$

Therefore, the explicit solution is $y=\sqrt{x^2{\tan }^{-1}x-x+{\tan }^{-1}x+9}$.