Question

In Problems 27-36solve the given initial-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{5}\mathit{y}\mathbf{=}\mathbf{35}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{,}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

$y\left(x\right)=-10e^{-2x}\cos x+9e^{-2x}\sin x+7e^{-4x}\text{.}$

Step-by-step solution

Complementary function

Consider the differential equation$y\text{'}\text{'}+4y\text{'}+5y=35e^{-4x},y\left(0\right)=-3,y\text{'}\left(0\right)=1$

The auxiliary equation corresponding to homogenous equation is:

$m^2+4m+5=0$

$$\begin{gathered} m=\frac{-4\pm \sqrt{4^2-4\left(1\right)\left(5\right)}}{2\left(1\right)}\text{By quadratic rule} \\ =\frac{-4\pm \sqrt{16-20}}{2} \\ =\frac{-4\pm 2i}{2} \\ =-2\pm i \end{gathered}$$

The roots of the auxiliary equation is$m_1=-2+i$and $m_2=-2-i$.

Therefore, the complementary function is.$y_c\left(x\right)=e^{-2x}\left(c_1\cos x+c_2\sin x\right)$

For particular solution

Let particular solution be$y_p\left(x\right)=A{e}^{-4x}$

Then,

$$\begin{gathered} y_p\text{'}\left(x\right)=-4A{e}^{-4x} \\ {y}_p\text{'}\text{'}\left(x\right)=16A{e}^{-4x} \end{gathered}$$

As$y_p\left(x\right)$is a solution of the equation $y\text{'}\text{'}+4y\text{'}+5y=35e^{-4x}$so,

$$\begin{gathered} 16A{e}^{-4x}+4\left(-4A{e}^{-4x}\right)+5A{e}^{-4x}=35e^{-4x} \\ 5A{e}^{-4x}=35e^{-4x} \\ 5A=35 \\ A=7 \end{gathered}$$

Thus, the particular solution is.$y_p\left(x\right)=7e^{-4x}$

Therefore, the general solution is

$$\begin{gathered} y\left(x\right)=y_c\left(x\right)+y_p\left(x\right) \\ =e^{-2x}\left(c_1\cos x+c_2\sin x\right)+7e^{-4x} \end{gathered}$$

Apply the first initial condition

To determine constants apply initial condition.

Put $y\left(0\right)=3$in$y\left(x\right)=e^{-2x}\left(c_1\cos x+c_2\sin x\right)+7e^{-4x}$

$$\begin{gathered} y\left(0\right)=e^0\left(c_1\cos 0+c_2\sin 0\right)+7e^0-3 \\ =c_1+0+7 \\ {c}_1=-10 \end{gathered}$$

Apply the second initial condition

Differentiate$y\left(x\right)$ with respect to x.

$y\text{'}\left(x\right)=-2e^{-2x}\left(c_1\cos x+c_2\sin x\right)+e^{-2x}\left(-c_1\sin x+c_2\cos x\right)-28e^{-4x}$

Put $y\text{'}\left(0\right)=1$in$y\text{'}\left(x\right)$

$$\begin{gathered} y\text{'}\left(0\right)=-2e^0\left(c_1\cos 0+c_2\sin 0\right)+e^0\left(-c_1\sin 0+c_2\cos 0\right)-28e^0 \\ 1=-2c_1+c_2-28 \\ 1=20+c_2-28\text{since}{c}_1=-10 \\ {c}_2=9 \end{gathered}$$

Now substituting values we have:

$$\begin{gathered} y\left(x\right)=e^{-2x}\left(c_1\cos x+c_2\sin x\right)+7e^{-4x} \\ y\left(x\right)=e^{-2x}(-10\cos x+9\sin x)+7e^{-4x}. \end{gathered}$$

Therefore, the general solution is $\mathit{y}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{10}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{+}\mathbf{9}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}\mathbf{7}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}$