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.

$\mathbf{(}\sqrt{x}\mathbf{+}\mathbf{x}\mathbf{)}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\sqrt{y}\mathbf{+}\mathbf{y}$

Short answer

The explicit solution is $y={\left(1-c_2{e}^{-\frac{x}{3}}\right)}^3$.

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$\frac{dy}{dx}=y^{\frac{2}{3}}-y$

Separate variables by cross multiplying$\frac{dy}{y^{\frac{2}{3}}-y}=dx$

Integrating we have:

$\int \frac{1}{{\mathrm{y}}^{\frac{2}{3}}-\mathrm{y}}\mathrm{dy}=\int \mathrm{d}\mathrm{x}$

Substitution

Consider the integral,

$\begin{array}{c}\int \frac{1}{y^{\frac{2}{3}}-y}dy=\int \frac{1}{y^{\frac{2}{3}}(1-y^{\frac{1}{3}})}dy\\ =\int \frac{y^{-\frac{2}{3}}}{(1-y^{\frac{1}{3}})}d\end{array}$

We will use $u$-substitution to solve the integral.

Let$u=1-y^{\frac{1}{3}}$

Then$du=-\frac{1}{3}{y}^{-\frac{2}{3}}dy$

Note that

$\begin{array}{c}-3du=-3\left(-\frac{1}{3}{y}^{-\frac{2}{3}}dx\right)\\ =y^{-\frac{2}{3}}dx\end{array}$

Integrating

So using the-substitution the integral becomes

$\begin{array}{c}\int \left(\frac{{\mathrm{y}}^{-2}}{1-{\mathrm{y}}^3}\right)=\int \frac{1}{\mathrm{y}}(-3)\mathrm{du}\\ =-3\int \frac{1}{\mathrm{y}}\mathrm{dy}\\ =-3\ln \left|\mathrm{u}\right|+\mathrm{e}\end{array}$

We substitute $u$to get the equation back in terms of $x$.

$\int \frac{y^{-\frac{2}{3}}}{(1-y^{\frac{1}{3}})}dy=-3\ln |1-y^{\frac{1}{3}}|+c$

Using this result we can solve equation (1).

$-3\ln |1-y^{\frac{1}{3}}|=x+c$

Find explicit solution

We solve for $y$in order to get the explicit solution.

role="math" localid="1667831091473" $\begin{array}{c}3\ln |1-y^{\frac{1}{3}}|=x+c\\ \ln |1-y^{\frac{1}{3}}|=-\frac{x}{3}+c_1\\ e^{\ln |1-y^{\frac{1}{3}}|}=e^{-\frac{x}{3}+c_1}\\ 1-y^{\frac{1}{3}}=c_2{e}^{-\frac{x}{3}}\\ y^{\frac{1}{3}}=1-c_2{e}^{-\frac{x}{3}}\\ y={\left(1-c_2{e}^{-\frac{x}{3}}\right)}^3\end{array}$

Therefore, the explicit solution is $y={\left(1-c_2{e}^{-\frac{x}{3}}\right)}^3$.