Question

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

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{y}\mathbf{-}{\mathbf{e}}^{\mathbf{3}\mathbf{x}}\mathbf{=}\mathbf{0}$

Short answer

The general solution to the given equation is $\mathrm{y}=\frac{{\mathrm{e}}^{3\mathrm{x}}}{2}+{\mathrm{Ce}}^{\mathrm{x}}$.

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 factorlocalid="1664262698741" role="math" $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$by the formula $\mathbf{\mu }\left(x\right)\mathbf{=}\mathbf{exp}\left[{\int }_{}^{}P\left(x\right)\mathrm{dx}\right]$.
• 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.

Solving the given equation

Given that,
$\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}-{\mathrm{e}}^{3\mathrm{x}}=0.......\left(1\right)$

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

$$\begin{gathered} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}-{\mathrm{e}}^{3\mathrm{x}}=0 \\ \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}={\mathrm{e}}^{3\mathrm{x}}.........\left(2\right) \end{gathered}$$

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

Where$\mathrm{P}(x)=-1$

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}-1\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\mathrm{x}}\end{array}$

Simplification method

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

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

Integrating both sides.

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

So, the solution is $\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{3}\mathbf{x}}}{\mathbf{2}}\mathbf{+}{\mathbf{Ce}}^{\mathbf{x}}\end{array}$.