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I have a step wrong while solving my IVP but I cannot find what. I will post my detailed solution in the hope someone sees where it goes awry: The IVP: $$t^2 \frac{dy}{dt}−t=1+y+ty,y(1)=4.$$

  1. I start by moving all function of y on the RHS: $$\displaystyle$$ $$\displaystyle t^2\frac{dy}{dt}-y-ty=1 +t$$ $$t^2\frac{dy}{dt}-(1+t)y=1 +t$$
  2. It is not in standard form so I continue by divinding by $t^2$: $$\frac{dy}{dt}-\frac{(1+t)}{t^2}y=\frac{1 +t}{t^2}$$
  3. With $h(t) = -\frac{(1+t)}{t^2}$ I can use integrating factor $e^{\int h(t)}$ = $e^{H(t)}$. With $H(t) = \frac{1}{t} - ln(t)$, I get :

$$e^{\frac{1}{t} - ln(t)}\frac{dy}{dt}- e^{\frac{1}{t} - ln(t)}\frac{(1+t)}{t^2}y=\frac{1 +t}{t^2}e^{\frac{1}{t} - ln(t)}$$ Which can be rewritten as: $$\frac{d}{dt}(e^{\frac{1}{t} - ln(t)} y) = \frac{1 +t}{t^2}e^{\frac{1}{t} - ln(t)}$$ $$\displaystyle e^{\frac{1}{t} - ln(t)} y = \int{ \frac{1 +t}{t^2}e^{\frac{1}{t} - ln(t)}} $$

$$\displaystyle e^{\frac{1}{t} - ln(t)} y = -e^{\frac{1}{t} - ln(t)} + C $$ 4. by multiplying with $e^-H(t)= e^{-\left(\frac{1}{t} - ln(t)\right)}$ we get:

$$ y(1) = 4 = -1 + Ce^{-\left(\frac{1}{t} - ln(t)\right)} \\ C = 5e$$

  1. The initial value condition gives that C:

$$y(1) = 4 = -1 + Ce^{-\left(\frac{1}{t} - ln(t)\right)} \\ C = 5e$$ $$y(t) = -1 + 5 e e^{-\left(\frac{1}{t} - ln(t)\right)} \\ y(t) = -1 + 5 e^{-\left(\frac{1}{t} - ln(t) - 1\right)}$$

But this appears to be wrong. Is there some step I am doing wrong?

EDIT: fixed -1 in exponent as per the comments.

EDIT 2: trying to plug in in the original equation

Simplifying we get: $$y(t) = -1 + 5 t e^{1-\frac{1}{t}}\\ y'(t) = 5 e^{1-\frac{1}{t}} + 5t \frac{1}{t^2}e^{1-\frac{1}{t}} \\ y'(t) = 5 e^{1-\frac{1}{t}} + 5 \frac{1}{t}e^{1-\frac{1}{t}}$$

$$t^2 \frac{dy}{dt}−t=1+y+ty \\ t^2 \left(5 e^{1-\frac{1}{t}} + 5t \frac{1}{t}e^{1-\frac{1}{t}} \right) -t = 1 -1 + 5 t e^{1-\frac{1}{t}} -t + 5 t^2 e^{1-\frac{1}{t}} $$

Seems everything is disappearing, but I still get wrong on the automatic assessment.

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4 Answers 4

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Isolate the derivative on the left side $$ t^2y'=1+t+y+yt=(1+t)(1+y). $$ This is a separable first order DE, so that $$ \ln(1+y)=\int\frac{dy}{1+y}=\int\frac{(1+t)dt}{t^2}=-\frac1t+\ln(t)+c \\~\\ (1+y)=Cte^{-1/t}. $$ Etc.

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  • $\begingroup$ Oh that was neat! $\endgroup$
    – Dovendyr
    Apr 14, 2021 at 12:32
  • $\begingroup$ Thank you everyone, I appreciated everyone's answer and I will vote @lutzlehmann up, since the answer was much more compact and a way to tackle the equation I did not thought about. $\endgroup$
    – Dovendyr
    Apr 15, 2021 at 7:03
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Your answer is correct note that $$e^{-1/t+\ln t}= te^{-1/t}.$$ Notr that $e^{ln z}=z$.

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Note Z Ahmed's answer; also, in the very last step, notice it is

$$5e^{-(\frac{1}{t}-\ln t-1)}.$$

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  • $\begingroup$ Right. thank you. Yet it is not right when I plug it back in the original equation. $\endgroup$
    – Dovendyr
    Apr 14, 2021 at 10:17
  • $\begingroup$ @Dovendyr Maybe I'm missing something but I think it should work out when you plug it in. Can you say how it's wrong? Like how much is it off by? $\endgroup$ Apr 14, 2021 at 10:21
  • $\begingroup$ Ok let me edit with pluggin in... just a sec :) $\endgroup$
    – Dovendyr
    Apr 14, 2021 at 10:21
  • $\begingroup$ Sorry it seems right. But the webwork still refuses my answer. $\endgroup$
    – Dovendyr
    Apr 14, 2021 at 10:30
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Note that:

\begin{align*} t^{2}\frac{dy}{dt} - t =1+y+ty \Leftrightarrow \frac{dy}{dt} = \frac{1}{t^{2}} + \frac{1}{t^2}y+\frac{1}{t}y+\frac{1}{t} = y(\frac{1}{t^2}+\frac{1}{t})+(\frac{1}{t^2}+\frac{1}{t}) \end{align*} Which is a linear equation. Solving this equation you conclude: \begin{equation*} y(t) = Ce^{-\frac{1}{t}}t - 1 \hspace{1cm}, C \in\mathbb{R} \end{equation*} Knowing that $y(1)=4$ you get: \begin{equation*} y(1) = 4 \Leftrightarrow C\frac{1}{e}-1 = 4 \Leftrightarrow C = 5e \end{equation*} Finally, you have: \begin{equation*} y(t) = 5e^{1-\frac{1}{t}}t-1 = 5e^{\frac{t-1}{t}}t-1 \end{equation*} Which verifies the inicial equation.

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