How to Solve and Verify Second Order Inhomogeneous ODEs?

[vj9]

Hello All,

I am stuck on the following question. Can you please help to find the solutions

Using the complementary function and particular integral method, find the solution of the diffential equation which satisfies y(0) = 1 and y'(0) = 0.

y'' + 3y' + 2y = 20cos2x

and then can you help about how to check the answer using technology.

 

[vin300]

D^2 + 3D +2=0
c1e^(-x) +c2e^(-2x) +(3sin2x -cos2x)/10
c1+c2-0.1=1
-c1-2c2 +6=0
c2=4.9, c1=-3.8

 

[vj9]

Can you be more specific . I tried to solve the way u suggested but i am stuck.

Many Thanks,
Vj9

 

[HallsofIvy]

First he found the characteristic equation for the homogeneous equation, $D^2+ 3D+ 2= (D+ 1)(D+ 2)= 0$ so that D= -1 and -2 and the "complementary solution" is $C_1e^{-x}+ C_2e^{-2x}$.

Since the right hand side is 20cos(2x), you look for a specific solution of the form Acos(2x)+ B sin(2x). Put that into the equation and solve for A and B.

 

[blue_raver22]

Hello,

To solve this second order inhomogeneous ODE, we first need to find the complementary function by setting the right side of the equation to 0. This gives us the characteristic equation: r^2 + 3r + 2 = 0. Solving this equation, we get r = -1 and r = -2. Therefore, the complementary function is yc = C1e^-x + C2e^-2x.

Next, we need to find the particular integral. Since the right side of the equation is a cosine function, we can assume a particular integral of the form yp = Acos2x + Bsin2x. Substituting this into the original equation, we get -4Acos2x - 4Bsin2x + 6Asin2x + 6Bcos2x + 2Acos2x + 2Bsin2x = 20cos2x. Simplifying, we get -2Acos2x + 4Bsin2x = 20cos2x. This gives us A = -10 and B = 0. Therefore, the particular integral is yp = -10cos2x.

The general solution is then given by y = yc + yp = C1e^-x + C2e^-2x - 10cos2x.

To check this solution using technology, you can use a graphing calculator or a software like MATLAB. Simply plug in the values of C1 and C2 and plot the function. Then, compare the graph with the given equation to see if they match. You can also plug in the initial conditions (y(0) = 1 and y'(0) = 0) to see if the solution satisfies them. Hope this helps!