MATH105 April 2017
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Question 04 (b)

Solve the following initial value problem:
 $y'=te^{t}y^{2},\quad y(0)=1$

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1

In order to solve this differential equation, you should use separation of variables. We can rewrite the differential equation as
${\frac {dy}{y^{2}}}=te^{t}dt$.
Integrate both sides of this equation and solve.

Hint 2

If you are struggling to compute
$\int te^{t}\,dt,$
try using integration by parts.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
 If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work.
 If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

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We will use separation of variables to solve the differential equation
${\frac {dy}{dt}}=te^{t}y^{2}$.
We rearrange this equation to get
${\frac {dy}{y^{2}}}=te^{t}dt$.
We are now ready to integrate both sides.
$\int {\frac {dy}{y^{2}}}=\int te^{t}\,dt.$
The integral on the left side is ${\frac {1}{y}}+C_{1}$, and the integral on the right side can be computed by integration by parts. Let $u=t$, $du=dt$, $dv=e^{t}\,dt$, $v=e^{t}.$ Applying integration by parts gives
$\int te^{t}\,dt=te^{t}\int e^{t}\,dt.$
This simplifies to
$\int te^{t}\,dt=te^{t}e^{t}+C_{2}$
so the differential equation has the general solution
${\frac {1}{y}}=te^{t}e^{t}+C$
We multiply both sides by 1:
$y^{1}=te^{t}+e^{t}C$.
Finally, we raise both sides to the power of 1.
$y=(te^{t}+e^{t}C)^{1}$.
It remains to solve the initial value problem. When t = 0, we should have y = 1. Thus
$1=(0e^{0}+e^{0}C)^{1},$
which simplifies to
$1=(1C)^{1}$.
This has the solution C = 0. So we get
$y=(te^{t}+e^{t})^{1}$.
Answer: $\color {blue}y=(te^{t}+e^{t})^{1}$


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