Chaya Galloway

2021-03-02

Solve differential equation $t{y}^{\prime }+\left(t+1\right)y=t,y\left(\mathrm{ln}2\right)=1$, t>0

Nathaniel Kramer

$t{y}^{\prime }+\left(t+1\right)y=t$
$dy/dx+\left(\left(t+1\right)/t\right)y=1$
where $p\left(t\right)=\left(\left(t+1\right)/t\right)$, q(t)=1
$I.F.={e}^{\int P\left(t\right)dt}$$={e}^{\int \left(\left(1+t\right)t\right)dt}$$=t{e}^{t}$
$t{e}^{t}y=\int t{e}^{t}dt$
$t{e}^{t}y=t\int {e}^{t}dt-\int \left[dt/dt\int {e}^{t}dt\right]dt$
$t{e}^{t}y=t{e}^{t}-\int \left[{e}^{t}\right]dt$
$t{e}^{t}y=t{e}^{t}-{e}^{t}+C$ (1)
$\left(\mathrm{ln}2\right){e}^{\mathrm{ln}2}\left(1\right)=\left(\mathrm{ln}2\right){e}^{\mathrm{ln}2}-{e}^{\mathrm{ln}2}+C$
$2\mathrm{ln}2=2\mathrm{ln}2-2+C$ (C=2)
$t{e}^{t}y=t{e}^{t}-{e}^{t}+2$
$y=\left(t{e}^{t}-{e}^{t}+2\right)/\left(t{e}^{t}\right)$
$y=\left(t{e}^{t}\right)/\left(t{e}^{t}\right)-{e}^{t}/\left(t{e}^{t}\right)+2/\left(t{e}^{t}\right)$
$y=1-1/t+2/t{e}^{t}$

nick1337

The differential equation is solved as follows:
$t{y}^{\prime }+\left(t+1\right)y=t$
To solve this equation, we use an integrating factor of ${e}^{\int \left(t+1\right)dt}={e}^{\frac{{t}^{2}}{2}+t}$
Multiplying both sides of the equation by the integrating factor gives:
$t{e}^{\frac{{t}^{2}}{2}+t}{y}^{\prime }+\left(t+1\right){e}^{\frac{{t}^{2}}{2}+t}y=t{e}^{\frac{{t}^{2}}{2}+t}$
The left side can be simplified using the product rule:
$\frac{d}{dt}\left(t{e}^{\frac{{t}^{2}}{2}+t}y\right)=t{e}^{\frac{{t}^{2}}{2}+t}$
Integrating both sides with respect to $t$ gives:
$t{e}^{\frac{{t}^{2}}{2}+t}y=\int t{e}^{\frac{{t}^{2}}{2}+t}dt$
The integral on the right side can be evaluated using integration by parts:
$u=t\phantom{\rule{1em}{0ex}}dv={e}^{\frac{{t}^{2}}{2}+t}dt$
$du=dt\phantom{\rule{1em}{0ex}}v=\int {e}^{\frac{{t}^{2}}{2}+t}dt$
Using the Gaussian integral, we find:
$v={e}^{\frac{{t}^{2}}{2}+t}$
Substituting back into the equation, we have:
$t{e}^{\frac{{t}^{2}}{2}+t}y={e}^{\frac{{t}^{2}}{2}+t}-\int {e}^{\frac{{t}^{2}}{2}+t}dt$
$t{e}^{\frac{{t}^{2}}{2}+t}y={e}^{\frac{{t}^{2}}{2}+t}-{e}^{\frac{{t}^{2}}{2}+t}+C$
$t{e}^{\frac{{t}^{2}}{2}+t}y=C$
Finally, solving for $y$ gives:
$y=\frac{C}{t{e}^{\frac{{t}^{2}}{2}+t}}$
To find the particular solution, we use the initial condition $y\left(\mathrm{ln}2\right)=1$. Substituting the values, we get:
$1=\frac{C}{\left(\mathrm{ln}2\right){e}^{\frac{\left(\mathrm{ln}2{\right)}^{2}}{2}+\mathrm{ln}2}}$
Simplifying this equation will give the specific value for $C$.

RizerMix

Given:
$t{y}^{\prime }+\left(t+1\right)y=t$
This is a first-order linear ordinary differential equation. To solve it, we can use an integrating factor. The integrating factor $I\left(t\right)$ is defined as:
$I\left(t\right)={e}^{\int \left(t+1\right)dt}={e}^{\left({t}^{2}/2+t\right)}$
Multiplying both sides of the equation by the integrating factor, we get:
${e}^{\left({t}^{2}/2+t\right)}t{y}^{\prime }+{e}^{\left({t}^{2}/2+t\right)}\left(t+1\right)y=t{e}^{\left({t}^{2}/2+t\right)}$
Now, we can simplify the left side of the equation using the product rule of differentiation:
$\frac{d}{dt}\left({e}^{\left({t}^{2}/2+t\right)}ty\right)=t{e}^{\left({t}^{2}/2+t\right)}$
Integrating both sides with respect to $t$, we have:
$\int \frac{d}{dt}\left({e}^{\left({t}^{2}/2+t\right)}ty\right)dt=\int t{e}^{\left({t}^{2}/2+t\right)}dt$
Simplifying the integrals, we get:
${e}^{\left({t}^{2}/2+t\right)}ty=\frac{1}{2}{e}^{\left({t}^{2}/2+t\right)}+C$
Dividing both sides by ${e}^{\left({t}^{2}/2+t\right)}t$, we obtain:
$y=\frac{1}{2t}+\frac{C}{t}{e}^{-\left({t}^{2}/2+t\right)}$
Now, we can use the initial condition $y\left(\mathrm{ln}2\right)=1$ to find the value of the constant $C$. Substituting $t=\mathrm{ln}2$ and $y=1$ into the equation, we have:
$1=\frac{1}{2\mathrm{ln}2}+\frac{C}{\mathrm{ln}2}{e}^{-\left({\mathrm{ln}}^{2}2/2+\mathrm{ln}2\right)}$
Simplifying this equation, we can solve for $C$:
$C=\left(1-\frac{1}{2\mathrm{ln}2}\right){e}^{\left({\mathrm{ln}}^{2}2/2+\mathrm{ln}2\right)}$
Substituting this value of $C$ back into the solution equation, we get the final solution:
$y=\frac{1}{2t}+\frac{\left(1-\frac{1}{2\mathrm{ln}2}\right){e}^{\left({\mathrm{ln}}^{2}2/2+\mathrm{ln}2\right)}}{t}{e}^{-\left({t}^{2}/2+t\right)}$

Vasquez

$y=\frac{2\mathrm{ln}|t+2|+\mathrm{ln}2-2\mathrm{ln}|\mathrm{ln}2+2|}{t{e}^{\frac{{t}^{2}}{2}+t}}.$
Explanation:
First, we rewrite the equation in standard form:
${\text{ty}}^{\prime }+\left(t+1\right)y=t.$
The integrating factor $I\left(t\right)$ is given by $I\left(t\right)={e}^{\int \left(t+1\right)\phantom{\rule{0.167em}{0ex}}dt}={e}^{\frac{{t}^{2}}{2}+t}$.
Multiplying both sides of the equation by the integrating factor, we get:
${\text{ty}}^{\prime }{e}^{\frac{{t}^{2}}{2}+t}+\left(t+1\right)y{e}^{\frac{{t}^{2}}{2}+t}=t{e}^{\frac{{t}^{2}}{2}+t}.$
Using the product rule for differentiation, we can simplify the left-hand side:
${\left(ty{e}^{\frac{{t}^{2}}{2}+t}\right)}^{\prime }=t{e}^{\frac{{t}^{2}}{2}+t}.$
Integrating both sides with respect to $t$, we obtain:
$ty{e}^{\frac{{t}^{2}}{2}+t}=\int t{e}^{\frac{{t}^{2}}{2}+t}\phantom{\rule{0.167em}{0ex}}dt.$
To solve the integral on the right-hand side, we use a change of variables. Let $u=\frac{{t}^{2}}{2}+t$. Then $du=\left(\frac{t}{2}+1\right)\phantom{\rule{0.167em}{0ex}}dt$ and $dt=\frac{2}{t+2}\phantom{\rule{0.167em}{0ex}}du$. Substituting these values, we have:
$\int t{e}^{\frac{{t}^{2}}{2}+t}\phantom{\rule{0.167em}{0ex}}dt=\int {e}^{u}\left(\frac{2}{t+2}\right)\phantom{\rule{0.167em}{0ex}}du=2\int {e}^{u}\left(\frac{1}{t+2}\right)\phantom{\rule{0.167em}{0ex}}du.$
Next, we integrate with respect to $u$:
$2\int {e}^{u}\left(\frac{1}{t+2}\right)\phantom{\rule{0.167em}{0ex}}du=2\mathrm{ln}|t+2|+C,$
where $C$ is the constant of integration.
Substituting this result back into the equation, we have:
$ty{e}^{\frac{{t}^{2}}{2}+t}=2\mathrm{ln}|t+2|+C.$
Now, we can solve for $y$:
$y=\frac{2\mathrm{ln}|t+2|+C}{t{e}^{\frac{{t}^{2}}{2}+t}}.$
Finally, using the initial condition $y\left(\mathrm{ln}2\right)=1$, we substitute $t=\mathrm{ln}2$ and $y=1$ into the equation to find the value of the constant $C$:
$1=\frac{2\mathrm{ln}|\mathrm{ln}2+2|+C}{\mathrm{ln}2{e}^{\frac{{\mathrm{ln}}^{2}2}{2}+\mathrm{ln}2}}.$
Solving for $C$, we have:
$C=\mathrm{ln}2-2\mathrm{ln}|\mathrm{ln}2+2|.$
Therefore, the solution to the given differential equation is:
$y=\frac{2\mathrm{ln}|t+2|+\mathrm{ln}2-2\mathrm{ln}|\mathrm{ln}2+2|}{t{e}^{\frac{{t}^{2}}{2}+t}}.$

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