Question

Question: In Problems 7-16, obtain the general solution to the equation.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}{\mathbf{x}}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{-}\mathbf{4}\mathbf{y}$

Short answer

The general solution to the given equation is$\mathrm{y}=\frac{{\mathrm{x}}^3}{3}{\mathrm{e}}^{-4\mathrm{x}}+{\mathrm{Ce}}^{-4\mathrm{x}}$

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathbf{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$. Here C is an arbitrary constant.

 Step 2: Solve the given equation

Given that,

$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2{\mathrm{e}}^{-4\mathrm{x}}-4\mathrm{y}.......\left(1\right)$

Write the equation (1) into standard form of linear equation.

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{x}}^2{\mathrm{e}}^{-4\mathrm{x}}-4\mathrm{y} \\ \frac{\mathrm{dy}}{\mathrm{dx}}+4\mathrm{y}={\mathrm{x}}^2{\mathrm{e}}^{-4\mathrm{x}}..........\left(2\right) \end{gathered}$$

Calculate the integrating factor of $\mathrm{\mu }\left(\mathrm{x}\right)$.

Where $\mathrm{P}(x)=4$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}4\mathrm{dx}}\\ & =& {\mathrm{e}}^{4\mathrm{x}}\end{array}$

 Step 3: Simplification method

Multiply $\mu \left(x\right)$in equation (2)

$\begin{array}{rcl}{e}^{4x}\frac{\mathrm{dy}}{\mathrm{dx}}+4e^{4x}\mathrm{y}& =& {\mathrm{x}}^2{\mathrm{e}}^{4\mathrm{x}}{\mathrm{e}}^{-4\mathrm{x}}\\ {\mathrm{e}}^{4\mathrm{x}}\frac{\mathrm{dy}}{\mathrm{dx}}+4{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}& =& {\mathrm{x}}^2\\ \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}\right]& =& {\mathrm{x}}^2\\ & & \end{array}$

Integrating both sides.

role="math" localid="1664260665464" $\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{y}\right]dx& =& {\int }_{}^{}{\mathrm{x}}^2\mathrm{dx}\\ {\mathrm{e}}^{4\mathrm{x}}\mathrm{y}& =& \frac{{\mathrm{x}}^3}{3}+\mathrm{C}\\ \mathrm{y}& =& \frac{{\mathrm{x}}^3}{3}{\mathrm{e}}^{-4\mathrm{x}}+{\mathrm{Ce}}^{-4\mathrm{x}}\end{array}$

So, the solution is$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \frac{{\mathbf{x}}^{\mathbf{3}}}{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\end{array}$.