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 $y={[-1+c(1+\sqrt{x}\left)\right]}^2$.

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

Separate variables we have:

$\begin{array}{c}(\sqrt{\mathrm{x}}+\mathrm{x})\mathrm{dy}=(\sqrt{\mathrm{y}}+\mathrm{y})\mathrm{dx}\\ \frac{1}{\sqrt{\mathrm{y}}+\mathrm{y}}\mathrm{dy}=\frac{1}{\sqrt{\mathrm{x}}+\mathrm{x}}\mathrm{dx}\end{array}$

Now we can integrate the left hand side in terms ofand the right hand side in terms of.

$\int \frac{1}{\sqrt{\mathrm{y}}+\mathrm{y}}\mathrm{dy}=\int \frac{1}{\sqrt{\mathrm{x}}+\mathrm{x}}\mathrm{dx}$

Evaluate ∫1y+ydyintegral

$\begin{array}{l}\sqrt{\text{y}}\text{=ufortheleftintegral.}\\ {\text{Then,y=u}}^{\text{2}}\text{anddy=2udu.}\\ \text{So,theintegralthenchangesto}\int \frac{\mathrm{dy}}{\sqrt{\mathrm{y}}+\mathrm{y}}=\int \frac{2\mathrm{u}}{\mathrm{u}+{\mathrm{u}}^2}\mathrm{du}\end{array}$

$\begin{array}{l}=2\int \frac{1}{1+\mathrm{u}}\mathrm{du}\\ =2\ln |1+\mathrm{u}|+{\mathrm{lnc}}_1\\ =2\ln |1+\sqrt{\mathrm{y}}|+{\mathrm{lnc}}_1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}substituteback}\sqrt{\mathrm{y}}=\mathrm{u}=2\ln (1+\sqrt{\mathrm{y}})+{\mathrm{lnc}}_1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}because}1+\sqrt{\mathrm{y}}>0\end{array}$

Similarly value of integral$\int \frac{dx}{\sqrt{x}+x}=2\ln (1+\sqrt{x})+\ln c_2$

Solution

Substituting these integral values back in the equation gives

$\begin{array}{c}2\ln (1+\sqrt{\mathrm{y}})+{\mathrm{lnc}}_1=2\ln (1+\sqrt{\mathrm{x}})+{\mathrm{lnc}}_2\\ \ln (1+\sqrt{\mathrm{y}})=\ln (1+\sqrt{\mathrm{x}})+\mathrm{lnc}\\ =\ln \left[\mathrm{c}\right(1+\sqrt{\mathrm{x}}\left)\right]1+\sqrt{\mathrm{y}}\\ =\mathrm{c}(1+\sqrt{\mathrm{x}})\sqrt{\mathrm{y}}\\ =-1+\mathrm{c}(1+\sqrt{\mathrm{x}})\\ \mathrm{y}={[-1+\mathrm{c}(1+\sqrt{\mathrm{x}}\left)\right]}^2\end{array}$