Science:Math Exam Resources/Courses/MATH101/April 2007/Question 04 (b)

MATH101 April 2007

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[hide] Question 04 (b)
Full-Solution Problem. Justify your answers and show all your work. Simplification of answers is not required. Solve the initial-value problem $y''-4y'+5y=5x^{2}-3x-2$ subject to $y(0)=0$ and $y'(0)=2$ .

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.

[show] Hint 1
First, let's find the particular solution. Solve the equation by noting that since the right hand side is degree 2 and the left hand side has a second derivative, then a solution is given by a quadratic polynomial of the form $\displaystyle y=Ax^{2}+Bx+C$

[show] Hint 2
Next, look for the homogeneous solution, i.e. the solution of the equation $\displaystyle y''-4y'+5y=0$ Adding solutions to this equation to the solution from hint 1 will also be solutions to the original differential equation. In fact, only the sum of the particular and the homogeneous solutions compose the full set of solutions.

[show] Hint 3
In your last step, use the initial conditions to solve for the constants in the homogeneous solution.

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.
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[show] Solution
Particular Solution We proceed as in the hints. First, we find a solution to $\displaystyle y''-4y'+5y=5x^{2}-3x-2$ by noticing that since the right hand side is a quadratic polynomial and the left hand side is a second order differential equation, we have that a solution is given by $\displaystyle y=Ax^{2}+Bx+C$ and we solve for A,B and C. Direct substitution yields ${\begin{aligned}5x^{2}-3x-2&=y''-4y'+5y\\&=2A-4(2Ax+B)+5(Ax^{2}+Bx+C)\\&=2A-8Ax-4B+5Ax^{2}+5Bx+5C\\&=5Ax^{2}+(5B-8A)x+5C-4B+2A\\\end{aligned}}$ Isolating to one side yields $\displaystyle (5-5A)x^{2}+(8A-5B-3)x+4B-2A-5C-2=0$ Since this must hold for all x values, we have that each of the individual coefficients must be zero. This gives ${\begin{aligned}5-5A&=0\\8A-5B-3&=0\\4B-2A-5C-2&=0\end{aligned}}$ Solving gives A=1, B=1 and C= 0. Hence, a solution to our differential equation is $\displaystyle y=x^{2}+x$ Homogeneous solution The story is not over yet. We could add the particular solution to solutions found via $\displaystyle y''-4y'+5y=0$ Adding these solutions together would give the full set of solutions. To solve the homogeneous part notice that the corresponding polynomial given by $\displaystyle r^{2}-4r+5=0$ has two imaginary roots given by (using the quadratic formula) $r={\frac {-(-4)\pm {\sqrt {(-4)^{2}-4(1)(5)}}}{2}}={\frac {4\pm {\sqrt {-4}}}{2}}=2\pm i$ Hence, to this new differential equation, we have the solutions $\displaystyle y=C_{1}e^{2x}\cos(x)+C_{2}e^{2x}\sin(x)$ Combining solutions, we have that $\displaystyle y=C_{1}e^{2x}\cos(x)+C_{2}e^{2x}\sin(x)+x^{2}+x$ . Initial Conditions Now we would be done except, this is an initial value problem so we need to solve for the constants. The initial conditions are $y(0)=0$ and $y'(0)=2$ . Hence, plugging in these values, we get that $\displaystyle 0=y(0)=C_{1}e^{2(0)}\cos(0)+C_{2}e^{2(0)}\sin(0)+(0)^{2}+(0)=C_{1}$ and that (taking the derivative via the product rule) ${\begin{aligned}y'&=C_{1}(2e^{2x}\cos(x)-\sin(x)e^{2x})+C_{2}(2e^{2x}\sin(x)+\cos(x)e^{2x})+2x+1\\2=y'(0)&=C_{1}(2e^{2(0)}\cos(0)-\sin(0)e^{2(0)})+C_{2}(2e^{2(0)}\sin(0)+\cos(0)e^{2(0)})+2(0)+1\\2=y'(0)&=0(2-0)+C_{2}(0+1)+2(0)+1\\1&=C_{2}\\\end{aligned}}$ NOTE: Above plugging in 0 for $C_{1}$ immediately would make the above computation simpler. Hence, the final solution is $\displaystyle y=e^{2x}\sin(x)+x^{2}+x$ . Phew!