I am trying to get through an assignment for my ODE class and was sick for a few lectures so I missed this part of the material.
The question asks me: Find the solution to the boundary value problem $$\frac{d^2y}{dt^2}-12\frac{dy}{dt}+35y=0, y(0)=4, y(1)=7$$
I'm not sure what the WebworK wants for a solution, because it won't accept the character 'x' in the answer box.
I've tried changing y'' and y' to r's (r^2 and r), and solve for the roots, 5 and 7. From there I'm stuck.
Any help is much appreciated. Thanks in advance!
You have that $y''-12y'+35y=0$. Let $y=e^{mx}$. Then, $$\left[m^2-12m+35\right]e^{mx}=0 \implies m^2-12m+35=0 \implies (m-7)(m-5)=0$$ Then, $m_1=5, m_2=7$. The basis for your null space is thus $\left[e^{5x},e^{7x}\right]$. Now, your general solution is $$y(x)=c_1e^{5x}+c_2e^{7x}$$ You must find $c_1$ and $c_2$, using the boundary conditions given. Thus, $$4=c_1+c_2$$ $$7=c_1e^5+c_2e^7$$
Solve the system of equations to find your constant values, plug them back in, and you have found your solution.
We have that $c_2=4-c_1$. Then, $7=c_1e^5+(4-c_1)e^7=c_1e^5+4e^7-c_1e^7 \implies c_1=\frac{7-4e^7}{e^5-e^7}$. So, $c_2=4-c_1=4-\frac{7-4e^7}{e^5-e^7}$.
The basis depends on the nature of your discriminant, that is, the $b^2-4ac$ in the quadratic formula. If $D>0$, then the basis is $[e^{m_1x},e^{m_2x}]$. If $D=0$, then you have $[e^{-\frac{b}{2a}x},xe^{-\frac{b}{2a}x}]$. If $D<0$, and $m_{1,2}=\alpha \pm \beta i$, then the basis is $[e^{\alpha x} \cos(\beta x),e^{\alpha x} \sin(\beta x)]$. The derivations require reduction of order, and for the last one, Euler's identity, and the change of basis formula. Note that this applies only to equations of the form $ay''+by'+cy=0$, where $a,b,c$ are arbitrary constants.