Question

In Problems 1-4 the given family of functions is the general solution of the differential equation on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

$$\begin{gathered} \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}\mathbf{,}\left(-\infty ,\infty \right) \\ \mathbf{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{y}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{2} \end{gathered}$$

Short answer

$\mathrm{y}=\frac{3}{5}{\mathrm{e}}^{4\mathrm{x}}-\frac{2}{5}{\mathrm{e}}^{-\mathrm{x}}$

Step-by-step solution

Find the value of y

We have the general solution of the differential equation $\mathrm{y}\text{'}\text{'}-3\mathrm{y}\text{'}-4\mathrm{y}=0$on an indicated interval as $\mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{4\mathrm{x}}+{\mathrm{c}}_2,\left(-\infty ,\infty \right)$

With the initial conditions

$\mathrm{y}\left(0\right)=1,\mathrm{y}\text{'}\left(0\right)=2$

Now we have to obtain the member of family at these conditions of following

First we have to apply point $\left(\mathrm{x},\mathrm{y}\right)=\left(0,0\right)$into the solution in equation (1), then we have

$$\begin{gathered} 1={\mathrm{c}}_1{\mathrm{e}}^0+{\mathrm{c}}_2{\mathrm{e}}^0 \\ {\mathrm{c}}_1+{\mathrm{c}}_2=1 \end{gathered}$$

First derivative of given equation,

$\mathrm{y}\text{'}=4{\mathrm{c}}_1{\mathrm{e}}^{4\mathrm{x}}-{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{x}}$

Applying Initial condition,

$$\begin{gathered} 2=4{\mathrm{c}}_1{\mathrm{e}}^0-{\mathrm{c}}_2{\mathrm{e}}^0 \\ 4{\mathrm{c}}_1-{\mathrm{c}}_2=2 \end{gathered}$$

By solving,

${\mathrm{c}}_1=\frac{3}{5}\mathrm{and}{\mathrm{c}}_2=\frac{2}{5}$

Substitute the value of ${\mathrm{c}}_1$and ${\mathrm{c}}_2$into the general solution of given D.E $\mathrm{y}\text{'}\text{'}-3\mathrm{y}\text{'}-4\mathrm{y}=0$by the way we can get

role="math" localid="1667912459535" $\mathrm{y}=\frac{3}{5}{\mathrm{e}}^{4\mathrm{x}}-\frac{2}{5}{\mathrm{e}}^{-\mathrm{x}}$

Hence the solution is$\mathbf{y}\mathbf{=}\frac{\mathbf{3}}{\mathbf{5}}{\mathbf{e}}^{\mathbf{4}\mathbf{x}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{5}}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}$