Question

In Problems 29-36 solve the given initial-value problem.

$$\begin{gathered} \mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{5} \end{gathered}$$

Short answer

$y=-\frac{7}{4}{e}^{-\frac{1}{2}x}+=\frac{11}{4}{e}^{\frac{3}{2}x}$

Step-by-step solution

Step 1: Determine the initial value of differential equation

Given,$4y\text{'}\text{'}-4y\text{'}-3y=0$

Initial conditions:

$y\left(0\right)=1 and y\text{'}\left(0\right)=5$

Consider$y=e^{mx}$as the solution of the differential equation,

Substitute$y=e^{mx},y\text{'}=m{e}^{mx}$ and$y\text{'}\text{'}=m^2{e}^{mx}$ into$4y\text{'}\text{'}-4y\text{'}-3y=0$ to obtain the auxiliary equation.

$$\begin{gathered} 4m^2{e}^{mx}-4m{e}^{mx}-3e^{mx}=0 \\ {e}^{mt}\left(4m^2-4m-3\right)=0 \end{gathered}$$

For any real$x,e^{mx}\ne 0$, we have

$$\begin{gathered} 4m^2-4m-3m=0 \\ \left(2m+1\right)\left(2m-1\right)=0 \end{gathered}$$

Then the roots are$m_1=-\frac{1}{2} and m_2=\frac{3}{2}$

Find the general solution of the homogeneous equation

Then the general solution of this homogeneous equation,

$4y\text{'}\text{'}-4y\text{'}-3y=0$

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

where $c_1\text{and}{c}_2$are an arbitrary constant.

Fourth, now we have to apply the given conditions in the general solution of our given differential equations as the following technique.

We have to apply with the point $\left(x,y\right)=\left(0,1\right)$ in the general solution (1) as

$1=c_1{e}^{\text{0}}+c_2{e}^{\text{0}}$

Then we have

$c_1+c_2=1·······\left(a\right)$

After that,take the first derivative for the general solution

The given differential equation as:

$y\text{'}=-\frac{1}{2}{c}_1{e^{\frac{-1}{2}x}}^{}+\frac{3}{2}{c}_2{e}^{\frac{3}{2}x}$

Then apply with the point as$\left(x,y\text{'}\right)=\left(0,5\right)$

$5=-\frac{1}{2}{c}_1{e}^0+\frac{3}{2}{c}_2{e^0}^{}$

Then we have

$-c_1+3c_2=10······\left(b\right)$

Then by solving the two equations

we find that

$c_1=-\frac{7}{4}\text{and} c_2=\frac{11}{4}$

Finally, substitute with the values of $c_1\text{and}{c}_2$general solution of our given differential equation,

Then$y=-\frac{7}{4}{e}^{-\frac{1}{2}x}+\frac{11}{4}{e}^{\frac{3}{2}x}$

Hence, the general solution is$\mathit{y}\mathbf{=}\mathbf{-}\frac{\mathbf{7}}{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}}\mathbf{+}\frac{\mathbf{11}}{\mathbf{4}}{\mathit{e}}^{\frac{\mathbf{3}}{\mathbf{2}}\mathbf{x}}$