Question

In Problems 35–44 use the Laplace transform to solve the giveninitial-value problem.

$\mathbf{35}\mathbf{.}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$\mathrm{y}\left(\mathrm{t}\right)=-1+{\mathrm{e}}^{\mathrm{t}}$

Step-by-step solution

To Find differential equation

Given

$\frac{\mathrm{dy}}{\mathrm{dt}}-\mathrm{y}=1,\mathrm{y}\left(0\right)=0$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{y}+1$

We first take the transform of each member of the differential equation.

$\mathrm{L}\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)=\mathrm{L}\left(\mathrm{y}\right)+\mathrm{L}\left(1\right)$

By substituting with : localid="1664104598735" role="math" $\mathrm{L}\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)=\mathrm{sy}\left(\mathrm{s}\right)-\mathrm{y}\left(0\right),\mathrm{y}\left(0\right)=0$

$\mathrm{sy}\left(\mathrm{s}\right)-\mathrm{y}\left(0\right)=\frac{1}{\mathrm{s}}+\mathrm{y}\left(\mathrm{s}\right)$

localid="1664103925753" $\mathrm{sy}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}}+\mathrm{y}\left(\mathrm{s}\right)$

localid="1664104654493" $(\mathrm{s}-1)\mathrm{y}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}}\mathrm{L}\left(\mathrm{y}\right)=\mathrm{y}\left(\mathrm{s}\right),\mathrm{La}\left(1\right)=\frac{1}{\mathrm{s}}$

Partial fraction

Solving the last equation for $\mathrm{y}\left(\mathrm{s}\right)$with partial fraction

$\mathrm{y}\left(\mathrm{s}\right)=\frac{1}{\mathrm{s}(\mathrm{s}-1)}=\frac{\mathrm{A}}{\mathrm{s}}+\frac{\mathrm{B}}{\mathrm{s}-1}$

$\mathrm{A}={\left(\frac{1}{\mathrm{s}-1}\right)}_{\mathrm{s}=0}=-1$

$\mathrm{B}={\left(\frac{1}{\mathrm{s}}\right)}_{\mathrm{s}=1}=1$

$\mathrm{y}\left(\mathrm{s}\right)=\frac{-1}{\mathrm{s}}+\frac{1}{\mathrm{s}-1}$

$\mathrm{y}\left(\mathrm{t}\right)=-1+{\mathrm{e}}^{\mathrm{t}}$Since${\mathrm{L}}^{-1}\left(\frac{1}{\mathrm{s}}\right)=1,{\mathrm{L}}^{-1}\left(\frac{1}{\mathrm{s}-1}\right)={\mathrm{e}}^{\mathrm{t}}$

Final proof

$\mathrm{y}\left(\mathrm{t}\right)=-1+{\mathrm{e}}^{\mathrm{t}}$