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.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\frac{{\mathbf{e}}^{\sqrt{x}}}{y}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{4}$

Short answer

The explicit solution is $y=2\sqrt{e^{\sqrt{x}}(\sqrt{x}-1)+4}$.

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$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{e}}^{\sqrt{\mathrm{x}}}}{\mathrm{y}}$

Separate variables by cross multiplying.

$\mathrm{ydy}={\mathrm{e}}^{\sqrt{\mathrm{x}}}\mathrm{dx}$

Now we integrate both sides:

$\int \mathrm{ydy}=\int {\mathrm{e}}^{\sqrt{\mathrm{x}}}\mathrm{dx}$

Substitution

Let u-substitution to solve the integral on the right hand side.

Let$y=\sqrt{x}$

Then$du=\frac{1}{2\sqrt{x}}dx$

Note that$2udu=dx$

So using the $u$substitution the integral becomes

$\int e^{\sqrt{x}}dx=\int 2u{e}^{u}du$

Integrate

In order to solve this integral we use integration by parts$\int \mathrm{wdv}=\mathrm{wv}-\int \mathrm{v}\mathrm{dx}$

The integral becomes

$\begin{array}{c}\int {\mathrm{e}}^{\sqrt{\mathrm{x}}}\mathrm{dx}=\int 2{\mathrm{ue}}^{\mathrm{u}}\mathrm{du}\\ =2{\mathrm{ue}}^{\mathrm{u}}-\int 2{\mathrm{e}}^{\mathrm{u}}\mathrm{du}\\ =2{\mathrm{ue}}^{\mathrm{u}}-2{\mathrm{e}}^{\mathrm{u}}+\mathrm{c}\end{array}$

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

$\begin{array}{c}\int {\mathrm{e}}^{\sqrt{\mathrm{x}}}\mathrm{dx}=2\sqrt{\mathrm{x}}{\mathrm{e}}^{\sqrt{\mathrm{x}}}-2{\mathrm{e}}^{\sqrt{\mathrm{x}}}+\mathrm{c}\\ =2{\mathrm{e}}^{\sqrt{\mathrm{x}}}(\sqrt{\mathrm{x}}-1)+\mathrm{c}\end{array}$

Using this result we can solve equation (1).

$\frac{1}{2}{y}^2=2e^{\sqrt{x}}(\sqrt{x}-1)+e$

Solve for y

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

$\begin{array}{c}\frac{1}{2}{y}^2=2e^{\sqrt{x}}(\sqrt{x}-1)+c\\ y^2=4e^{\sqrt{x}}(\sqrt{x}-1)+c_1\\ y=\pm 2\sqrt{e^{\sqrt{x}}(\sqrt{x}-1)+c_1}\end{array}$

Apply initial condition

The initial condition$y\left(1\right)=4$ means we need a positive-value so

$\mathrm{y}=2\sqrt{{\mathrm{e}}^{\sqrt{\mathrm{x}}}(\sqrt{\mathrm{x}}-1)+{\mathrm{c}}_1}$

Use the initial condition to solve for the constant:

$\begin{array}{l}4=2\sqrt{{\mathrm{e}}^{\sqrt{1}}(\sqrt{1}-1)+{\mathrm{c}}_1}\\ 2=\sqrt{\mathrm{e}\left(0\right)+{\mathrm{c}}_1}\\ 2=\sqrt{{\mathrm{c}}_1}\\ 4={\mathrm{c}}_1\end{array}$

Therefore, the explicit solution is$y=2\sqrt{e^{\sqrt{x}}(\sqrt{x}-1)+4}$