Question

In Problemssolve the given initial-value problem.

$\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{14}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{11}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$y=-\frac{1}{5}{e}^{-2x}+\frac{186}{5}{e}^{\frac{x}{2}}-7x^2-19x-37$

Step-by-step solution

Complementary function

Consider the differential equation$2y\text{'}\text{'}+3y\text{'}-2y=14x^2-4x-11,y\left(0\right)=0,y\text{'}\left(0\right)=0$

The auxiliary equation corresponding to homogenous equation is:$2m^2+3m-2=0$

we find $m_1=\frac{1}{2}$and$m_2=-2$

So, the complementary function is$y_c=C_1{e}^{\frac{x}{2}}+C_2{e}^{-2x}$

For particular solution

Let particular solution be$y_p=A{x}^2+Bx+C{y}_p$

Then substituting this into the given differential equation yields

$$\begin{gathered} 2\left(2A\right)+3(2Ax+B)-2\left(A{x}^2+Bx+C\right)=14x^2-4x-11 \\ -2A{x}^2+(6A-2B)x+(4A+3B-2C)=14x^2-4x-11 \end{gathered}$$

On equating the coefficients of like terms, we get

$$\begin{gathered} -2A=14 \\ 6A-2B=-4 \\ 4A+3B-2C=-11 \\ A=-7,B=-19\text{and}C=-37 \end{gathered}$$

So,

$y_p=-7x^2-19x-37$

Find the Solution 

We have$y=C_1{e}^{\frac{x}{2}}+C_2{e}^{-2x}-7x^2-19x-37$

Then$y\text{'}=\frac{1}{2}{C}_1{e}^{\frac{x}{2}}-2C_2{e}^{-2x}-14x-19$

Now$y\left(0\right)=0$implies$0=C_1+C_2-37$

And $y\text{'}\left(0\right)=0$implies$0=\frac{C_1}{2}-2C_2-19$

By solving, we get $C_2=-\frac{1}{5}$ and $C_1=\frac{186}{5}$

Thus, the solution of the given initial-value problem is

$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{5}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{+}\frac{\mathbf{186}}{\mathbf{5}}{\mathit{e}}^{\frac{\mathbf{x}}{\mathbf{2}}}\mathbf{-}\mathbf{7}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{19}\mathit{x}\mathbf{-}\mathbf{37}$