Question

Section 5.2 Problem 20:

Solve the initial value problem and graph the solution.


`36y'' - 12y' + y = 0`

`y(0) = 3`

`y'(0) = (5)/(2)`
​

Answer 1

`y(x)=3e^{(x)/(6)}+(12)/(7)xe^{(x)/(6)}` (See attached graph)

Explanation:

To solve a second-order homogeneous differential equation, we need to substitute each term with the auxiliary equation
`am^2+bm+c=0` where the values of
`m` are the roots:

`36y''-12y'+y=0\\\\36m^2-12m+1=0\\\\(6m-1)^2=0\\\\6m-1=0\\\\6m=1\\\\m=(1)/(6)`

Since the values of
`m` are equal real roots, then the general solution is
`y(x)=C_1e^(m_1x)+C_2xe^(m_1x)`.

Thus, the general solution for our given differential equation is
`y(x)=C_1e^{(x)/(6)}+C_2xe^{(x)/(6)}`.

To account for both initial conditions, take the derivative of
`y(x)`, thus,
`y'(x)=(C_1)/(6)e^{(x)/(6)}+(C_2)/(6)e^{(x)/(6)}+C_2e^{(x)/(6)}`

Now, we can create our system of equations given our initial conditions:

`y(x)=C_1e^{(x)/(6)}+(C_2)/(6)xe^{(x)/(6)}\\ \\y(0)=C_1e^{(0)/(6)}+(C_2)/(6)(0)e^{(0)/(6)}=3\\ \\C_1=3`

`y'(x)=(C_1)/(6)e^{(x)/(6)}+(C_2)/(6)e^{(x)/(6)}+C_2e^{(x)/(6)}\\\\y'(0)=(C_1)/(6)e^{(0)/(6)}+(C_2)/(6)e^{(0)/(6)}+C_2e^{(0)/(6)}=(5)/(2)\\ \\(C_1)/(6)+(C_2)/(6)+C_2=(5)/(2)`

We then solve the system of equations, which becomes easy since we already know that
`C_1=3`:

`(C_1)/(6)+(C_2)/(6)+C_2=(5)/(2)\\\\(3)/(6)+(C_2)/(6)+C_2=(5)/(2)\\\\3+C_2+6C_2=15\\\\7C_2=12\\\\C_2=(12)/(7)`

Thus, our final solution is:

`y(x)=C_1e^{(x)/(6)}+C_2xe^{(x)/(6)}\\\\y(x)=3e^{(x)/(6)}+(12)/(7)xe^{(x)/(6)}`