Science:Math Exam Resources/Courses/MATH101/April 2006/Question 04 (b)

MATH101 April 2006

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Question 04 (b)
Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. Solve the initial-value problem Solve the initial-value problem $\displaystyle y''+2y'=\cos(x)$ subject to the conditions $y(0)=1$ and $y'(0)=-1$

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
This is actually kind of a first order differential equation disguised as a second order differential equation. Try an integrating factor.

Hint 2
Multiply both sides of the equation by $e^{-2x}$ where the -2 above came from the coefficient of the $y'$ term.

Hint 3
Even after all this work the right hand side integral is very tedious. It's an example of the cyclic type so try using integration by parts twice and noticing that you end up where you started.

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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We proceed as in the hint. Multiplying both sides by $e^{2x}$ gives $\displaystyle e^{2x}y''+2e^{2x}y'=e^{2x}\cos x$ Now, the left hand side is just $(e^{2x}y')'$ and so $\displaystyle (e^{2x}y')'=e^{2x}\cos x$ Integrating both side, we see that the integral of the right hand side above is difficult - so we compute it separately. We proceed by integration by parts. Let ${\begin{aligned}u=e^{2x}\quad &\quad v=\sin x\\du=2e^{2x}dx\quad &\quad dv=\cos xdx\\\end{aligned}}$ and so the integral becomes ${\begin{aligned}\int e^{2x}\cos x\,dx&=e^{2x}\sin x-2\int e^{2x}\sin x\,dx\end{aligned}}$ We use parts again. Let ${\begin{aligned}u=e^{2x}\quad &\quad v=-\cos x\\du=2e^{2x}dx\quad &\quad dv=\sin xdx\\\end{aligned}}$ Then, we have ${\begin{aligned}\int e^{2x}\cos x\,dx&=e^{2x}\sin x-2\int e^{2x}\sin x\,dx\\&=e^{2x}\sin x-2\left(-e^{2x}\cos x+2\int e^{2x}\cos x\,dx\right)\\&=e^{2x}\sin x+2e^{2x}\cos x-4\int e^{2x}\cos x\,dx\end{aligned}}$ This last integral on the right is the same as the integral on the left (up to a constant). Hence, we have ${\begin{aligned}\int e^{2x}\cos x\,dx+4\int e^{2x}\cos x\,dx&=e^{2x}\sin x+2e^{2x}\cos x+C\\5\int e^{2x}\cos x\,dx&=e^{2x}\sin x+2e^{2x}\cos x+C\\\int e^{2x}\cos x\,dx&={\frac {e^{2x}\sin x}{5}}+{\frac {2e^{2x}\cos x}{5}}+D\\\end{aligned}}$ where $D={\frac {C}{5}}$. After this gigantic diversion, we return to the point. We had $\displaystyle (e^{2x}y')'=e^{2x}\cos x$ and integrating both sides, we get $\displaystyle e^{2x}y'=\int e^{2x}\cos x\,dx={\frac {e^{2x}\sin x}{5}}+{\frac {2e^{2x}\cos x}{5}}+D$ Dividing both sides by $e^{2x}$, we have $\displaystyle y'={\frac {\sin x}{5}}+{\frac {2\cos x}{5}}+De^{-2x}$ Now, we solve for $D$. We plug in the initial condition $y'(0)=-1$ and see that $\displaystyle -1=y'(0)={\frac {\sin(0)}{5}}+{\frac {2\cos(0)}{5}}+De^{-2(0)}={\frac {2}{5}}+D$ and so isolating, we see that $D=-{\frac {7}{5}}$. Again integrating both sides of $y'$ gives ${\begin{aligned}y&=\int {\frac {\sin x}{5}}+{\frac {2\cos x}{5}}-{\frac {7}{5}}e^{-2x}\,dx\\&={\frac {-\cos x}{5}}+{\frac {2\sin x}{5}}+{\frac {7e^{-2x}}{10}}+E\end{aligned}}$ where E is another constant. To compute E, we use the other initial condition that $y(0)=1$ to see that ${\begin{aligned}1=y(0)&={\frac {-\cos(0)}{5}}+{\frac {2\sin(0)}{5}}+{\frac {7e^{-2(0)}}{10}}+E=-{\frac {1}{5}}+{\frac {7}{10}}+E\end{aligned}}$ and so $E={\frac {1}{2}}$. Combining gives ${\begin{aligned}y&={\frac {-\cos x}{5}}+{\frac {2\sin x}{5}}+{\frac {7e^{-2x}}{10}}+{\frac {1}{2}}\end{aligned}}$ and we are finished!

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