Question

Question: In Problems 7-16, obtain the general solution to the equation.

$\left(t+y+1\right)\mathbf{dt}\mathbf{-}\mathbf{dy}\mathbf{=}\mathbf{0}$

Short answer

The general solution to the given equation is $\mathrm{y}=-\mathrm{t}-2+{\mathrm{Ce}}^{-\mathrm{t}}$.

Step-by-step solution

Method for solving linear equations

• Write the equation in the standard form $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.
• Calculate the integrating factor by $\mathbf{\mu }\left(x\right)$the formula .
• Multiply the equation in standard form by$\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and, recalling that the left-hand side is just $\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}$, obtain

$$\begin{gathered} \mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathit{P}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{y}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\left[}\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{y}\mathbf{\right]}\mathbf{=}\mathit{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathit{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \end{gathered}$$

• Integrate the last equation and solve for y by dividing by $\mathbf{\mu }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$to obtain $\mathbf{y}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{[}{\mathbf{\int }}_{}^{}\mathbf{\pi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{Q}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{c}\mathbf{]}$. Here C is an arbitrary constant.

Solve the given equation

Given that,

$(\mathrm{t}+\mathrm{y}+1)\mathrm{dt}-\mathrm{dy}=0........\left(1\right)$

Write the equation (1) into standard form of linear equation.

$\begin{array}{rcl}(\mathrm{t}+\mathrm{y}+1)\mathrm{dt}-\mathrm{dy}& =& 0\\ -\mathrm{dy}& =& -(\mathrm{t}+\mathrm{y}+1)\mathrm{dt}\\ \frac{\mathrm{dy}}{\mathrm{dt}}& =& (\mathrm{t}+\mathrm{y}+1)\\ \frac{\mathrm{dy}}{\mathrm{dt}}-\mathrm{y}& =& \mathrm{t}+1...........\left(2\right)\end{array}$

Calculate the integrating factor of $\mathrm{\mu }\left(\mathrm{t}\right)$.

Where $\mathrm{P}\left(\mathrm{t}\right)=-1$.

Then,

$\begin{array}{rcl}\mathrm{\mu }\left(\mathrm{x}\right)& =& {\mathrm{e}}^{{\int }_{}^{}\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}}\\ & =& {\mathrm{e}}^{{\int }_{}^{}-1\mathrm{dx}}\\ & =& {\mathrm{e}}^{-\mathrm{t}}\end{array}$

Simplification method

Multiply $\mathrm{\mu }\left(\mathrm{t}\right)$in equation (2)

$\begin{array}{rcl}{\mathrm{e}}^{-\mathrm{t}}\frac{\mathrm{dy}}{\mathrm{dt}}-{\mathrm{e}}^{-\mathrm{t}}\mathrm{y}& =& {\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)\\ \frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{-\mathrm{t}}\mathrm{y}\right]& =& {\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)\end{array}$

Integrating both sides.

role="math" localid="1664255973738" $\begin{array}{rcl}{\int }_{}^{}\frac{\mathrm{d}}{\mathrm{dt}}\left[{\mathrm{e}}^{-\mathrm{t}}\mathrm{y}\right]dt& =& {\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)dt\\ {\mathrm{e}}^{-\mathrm{t}}\mathrm{y}& =& {\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{t}+1\right)\mathrm{dt}\\ & =& {\int }_{}^{}\left({\mathrm{e}}^{-\mathrm{t}}\mathrm{t}+{\mathrm{e}}^{-\mathrm{t}}\right)\mathrm{dt}\\ & =& {\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{tdt}+{\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\\ {\mathrm{e}}^{-\mathrm{t}}\mathrm{y}& =& {\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{tdt}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}\end{array}$

Find the value of role="math" localid="1664256337535" $\begin{array}{rcl}& & {\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{tdt}\end{array}$separately.

Let $\mathrm{u}=\mathrm{t},\mathrm{dv}={\mathrm{e}}^{-\mathrm{tdt}}$

$\mathrm{du}=\mathrm{dt},\mathrm{v}={\mathrm{e}}^{-\mathrm{t}}$

Then,

$\begin{array}{rcl}{\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{tdt}& =& -{\mathrm{te}}^{-\mathrm{t}}+{\int }_{}^{}{\mathrm{e}}^{-\mathrm{t}}\mathrm{dt}\\ & =& {\mathrm{te}}^{-\mathrm{t}}-{\mathrm{e}}^{-\mathrm{t}}\end{array}$

Substitute the founded value in equation (3).

$\begin{array}{rcl}{\mathrm{e}}^{-\mathrm{t}}\mathrm{y}& =& -t{e}^{-t}-e^{-t}-{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}\\ {\mathrm{e}}^{-\mathrm{t}}\mathrm{y}& =& -{\mathrm{te}}^{-\mathrm{t}}-2{\mathrm{e}}^{-\mathrm{t}}+\mathrm{C}\\ \mathrm{y}& =& -\mathrm{t}-2+{\mathrm{Ce}}^{\mathrm{t}}\end{array}$

So, the solution is$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{-}\mathbf{t}\mathbf{-}\mathbf{2}\mathbf{+}{\mathbf{Ce}}^{\mathbf{t}}\end{array}$