Question

In Problems 35–38, $\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{3}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathbf{-}\mathbf{2}\mathit{x}$is a two-parameter family of the second-order DE $\mathit{y}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{6}\mathit{x}\mathbf{+}\mathbf{4}$. Find a solution of the second-order IVP consisting of this differential equation and the given initial conditions.

$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y=\frac{1}{2}{e}^{3x}-\frac{1}{2}{e}^{-x}-2x$

Step-by-step solution

Definition of differential equation

A differential equation is defined as an equation that consists of the derivatives of one or more dependent functions with respect to one or more independent functions.

Take the derivative with respect to x and find solution of the second order IVP

Given: $y=c_1{e}^{3x}+c_2{e}^{-x}-2x······\left(1\right)$

Take the derivative with respect to x.

$y\text{'}=3c_1{e}^{3x}-c_2{e}^{-x}-2······\left(2\right)$

Initial condition in (1).

$\begin{array}{c}{c}_1+c_2=0\\ c_1=-c_2\end{array}$

Initial condition in (2) and substitute $c_1=-c_2$, we get

$\begin{array}{c}3c_1-c_2-2=0\\ 3c_1+c_1=2\\ 4c_1=2\\ c_1=\frac{1}{2}\end{array}$

Substitute $c_1=\frac{1}{2}$,

So, $c_2=-\frac{1}{2}$

Substitute the values of $c_1$and $c_2$in the given equation we get,

$y=\frac{1}{2}{e}^{3x}-\frac{1}{2}{e}^{-x}-2x$

Therefore, the solution of the second order IVP consisting of this differential equation and the given initial conditions is $y=\frac{1}{2}{e}^{3x}-\frac{1}{2}{e}^{-x}-2x$.