Question

Solve the following IVP

y"-10y'+9y=5t;y(0)=-1,y'(0)=2

Answer 1

Given the differential equation,

$$y''-10y'+9y=5t$$

such that

$$y(0)=-1,y'(0)=2$$

The homogenous part of the equation is:

$$y''-10y'+9y=0$$

Let $m$ be the root of the auxiliary equation such that:

$$m^2-10m+9=0\\ (m-9)(m-1)=0\\ m =9 \text{ and } m=1$$

are the roots of the equation.

Therefore the particular solution is:

$$y_c = c_1e^t + c_2e^{9t}$$

We determine the particular solution to the non-homogenous part of the given differential equation using the method of undetermined coefficient.

$$y''-10y'+9y=5t$$

The solution is of the form:

$$y_p = a_1+a_2t$$

Solving for the unknown constants $a_1$ and $a_2$ :

$$y_p' = a_2\\ y_p'' = 0\\$$

Substituting the particular solution into the differential equation, we get:

$$-10a_2+9(a_1+a_2t) =5t\\$$

Simplifying further, we have:

$$9a_1-10a_2+9a_2t = 5t$$

By equating coefficients:

$$9a_1-10a_2=0\\ 9a_2t = 5t$$

Solving for the system yields:

$$a_1 = \frac{50}{81}\\ a_2 = \frac{5}{9}$$

Substituting $a_1 \text{ and } a_2 \text{ into } y_p, \text{ we have }:$

$$y_p = \frac{5t}{9}+\frac{50}{81}$$

Thus, the general solution is:

$$y= y_c+y_p = c_1e^t + c_2e^{9t}+ \frac{5t}{9}+\frac{50}{81}$$

We now solve for the unknown constants using the initial conditions:

$$\dfrac{dy}{dt} = c_1e^t+9c_2e^{9t}+\frac{5}{9}$$

Substituting y(0)=-1 into y (the general solution), we have:

$$c_1+c_2+\frac{50}{81}=-1$$

Substitute y'(0) = 2 into $\frac{dy}{dt}$ we have:

$$c_1+9c_2+\frac{5}{9}=2$$

Solving the system gives:

$$c_1 = -2\\ c_2 = \frac{31}{38}$$

Substitute the values of $c_1 \text{ and } c_2$ into the general solution, we have:

$$y= \frac{1}{81}(31e^{9t}-162e^t+45t+50)$$

which is the required solution.