Question

In Problemssolve the given initial-value problem.

$\mathbf{5}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{-}\mathbf{6}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{10}$

Short answer

$y=-200+200e^{-\frac{x}{5}}-3x^2+30x$

Step-by-step solution

Complementary function

consider the differential equation$5y\text{'}\text{'}+y\text{'}=-6x,y\left(0\right)=0,y\text{'}\left(0\right)=-10$

The auxiliary equation corresponding to homogenous equation is:

$5m^2+m=0$we find$m_1=-\frac{1}{5}$ and$m_2=0$

So, the complementary function is$y_c=C_1+C_2{e}^{-\frac{x}{5}}$

Particular solution

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

Then

$$\begin{gathered} y_p\text{'}=2Ax+B \\ =2A \end{gathered}$$

Now, substituting these into the given differential equation gives.

$10A+2Ax+B=-6x$

On equating the coefficients of like terms, we get

$$\begin{gathered} 2A=-6 \\ 10A+B=0 \end{gathered}$$

We have$A=-3,B=30$

So,$y_p=-3x^2+30x$

General solution

Thus, the general solution of the given differential equation is

$y=y_c+y_p$

$y=C_1+C_2{e}^{-\frac{x}{5}}-3x^2+30x$

$y\text{'}=-\frac{C_2}{5}{e}^{-\frac{x}{5}}-6x+30$

Put $y\left(0\right)=0$implies$0=C_1+C_2$

And$y\text{'}\left(0\right)=-10$gives$-10=-\frac{C_2}{5}+30$

$$\begin{gathered} -\frac{C_2}{5}=-40 \\ {C}_2=200 \end{gathered}$$

And$C_1=-200$

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

$\mathit{y}\mathbf{=}\mathbf{-}\mathbf{200}\mathbf{+}\mathbf{200}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{x}}{\mathbf{5}}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{30}\mathit{x}$