Question

In problem 7-16, solve the equation. $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\frac{t}{\mathbf{x}{\mathbf{e}}^{t+2x}}$

Short answer

The solution of the given differential equation is ${\mathbf{e}}^{2x}\left(2x-1\right)\mathbf{+}\mathbf{4}{\mathbf{e}}^{-t}\left(t+1\right)\mathbf{=}\mathbf{C}$.

Step-by-step solution

Concept of Separable Differential Equation

A first-order ordinary differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{f}\left(x,y\right)$is referred to as separable if the function in the right-hand side of the equation is expressed as a product of two functions $\mathbf{g}\left(x\right)$that is a function of x alone and $\mathbf{h}\left(y\right)$that is a function of y alone.

A separable differential equation can be expressed as $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{h}\left(y\right)$. By separating the variables, the equation follows $\frac{\mathrm{dy}}{\mathbf{h}\left(y\right)}\mathbf{=}\mathbf{g}\left(x\right)\mathbf{dx}$. Then, on direct integration of both sides, the solution of the differential equation is determined.

Solution of the Equation

The given equation is

$\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{t}}{\mathrm{x}{\mathrm{e}}^{\mathrm{t}+2\mathrm{x}}}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

After separating the variables, equation (1) can be written as

$\begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\mathrm{t}}{\mathrm{x}{\mathrm{e}}^{\mathrm{t}}{\mathrm{e}}^{2\mathrm{x}}}\\ \mathrm{x}{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

Integrate both sides of equation (2). It results,

$\begin{array}{c}\int \mathrm{x}{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=\int \mathrm{t}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\\ \frac{1}{2}\int 2\mathrm{x}{\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}=\left(-1\right)\int -\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\end{array}$

$\begin{array}{c}\frac{1}{2}\int \mathrm{x}\mathrm{d}\left({\mathrm{e}}^{2\mathrm{x}}\right)=\left(-1\right)\int \mathrm{t}\mathrm{d}\left({\mathrm{e}}^{-\mathrm{t}}\right)\\ \frac{1}{2}\left[\mathrm{x}{\mathrm{e}}^{2\mathrm{x}}-\int {\mathrm{e}}^{2\mathrm{x}}\mathrm{dx}\right]=\left(-1\right)\left[\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}-\int {\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\right]\left[\int \mathrm{u}\mathrm{dv}=\mathrm{uv}-\int \mathrm{v}\mathrm{du}\right]\\ \frac{1}{2}\left[\mathrm{x}{\mathrm{e}}^{2\mathrm{x}}-\frac{{\mathrm{e}}^{2\mathrm{x}}}{2}\right]=\left(-1\right)\left[\mathrm{t}{\mathrm{e}}^{-\mathrm{t}}+{\mathrm{e}}^{-\mathrm{t}}\right]+\mathrm{k}\left[\mathrm{k}=\mathrm{IntegratingConstant}\right]\\ {\mathrm{e}}^{2\mathrm{x}}\left(2\mathrm{x}-1\right)=-4{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)+4\mathrm{k}\\ {\mathrm{e}}^{2\mathrm{x}}\left(2\mathrm{x}-1\right)+4{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)=\mathrm{C}\left[\mathrm{C}=4\mathrm{k}=\mathrm{Constant}\right]\end{array}$

Therefore, the solution of the given equation is ${\mathbf{e}}^{2x}\left(2x-1\right)\mathbf{+}\mathbf{4}{\mathbf{e}}^{-t}\left(t+1\right)\mathbf{=}\mathbf{C}$.