I don't understand how to solve this. I can solve the left side and get yc, but I have no idea how to solve for the right side.
y''-3y'+2y=(e^2t)/(e^t+1)
http://www.wolframalpha.com/input/?i=Solve+y%27%27-3y%27%2B2y%5C:%3D%5C:%5Cfrac%7Be%5E%7B2t%5C:%7D%7D%7B%5Cleft(e%5E%7Bt%5C:%5C:%7D%2B1%5Cright)%7D
ok. I assume you can get the homogeneous solution.
Since x^2-3x+2=0 means x=1 or 2, that would be
y = c1 e^x + c2 e^(2x)
Now for the fun part. The right side indicates we might have success using undetermined coefficients, with something like
u = ln(e^x+1)
since the derivatives will produce powers of (e^x+1) in the denominator. It is kind of like partial fractions. You will wind up with
y = c1 e^x + c2 e^(2x) + xe^(2x) - e^x ln(e^x+1) - e^(2x) ln(e^x+1)
Thanks for the help, Steve. I didn't see that. I'll give it a shot.
To solve the given differential equation, y'' - 3y' + 2y = (e^(2t))/(e^t + 1), you can use the method of undetermined coefficients.
First, solve the homogeneous equation: y'' - 3y' + 2y = 0.
Assume a solution of the form y_h = e^(rt), where r is a constant.
Substituting this solution into the homogeneous equation, we get:
(r^2 - 3r + 2)e^(rt) = 0.
The characteristic equation is r^2 - 3r + 2 = 0.
Solve this quadratic equation to find the roots r1 and r2.
Once you have the roots, you can find the homogeneous solution y_h by using the formula:
y_h(t) = C1*e^(r1 * t) + C2*e^(r2 * t), where C1 and C2 are constants determined by the initial conditions.
Now, let's find the particular solution y_p.
Since the right side of the equation is (e^(2t))/(e^t + 1), we will try a particular solution of the form:
y_p = A*e^(2t)/(e^t + 1), where A is a constant.
Differentiate y_p:
y_p' = (2A*e^(2t)*(e^t + 1) - A*(e^(2t)*e^t))/(e^t + 1)^2
= (2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^2.
Differentiate again:
y_p'' = (2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^2 - 2(2A*e^(2t)*(e^t + 1) - A*e^(3t))*(e^t + 1)/((e^t + 1)^2)^2
= (2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^2 - 2(2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^3.
Substituting y_p, y_p', and y_p'' into the original equation:
[(2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^2 - 3(2A*e^(2t)*(e^t + 1) - A*e^(3t))/(e^t + 1)^3
+ 2(A*e^(2t)/(e^t + 1))] = (e^(2t))/(e^t + 1).
Expand and simplify the equation, collecting like terms:
[(-2A + 3A - 2A)*e^(3t) + (4A - 2A - 4A)*e^(2t)]/(e^t + 1)^3 = (e^(2t))/(e^t + 1).
Finally, set the coefficients of e^(3t) and e^(2t) on the left side equal to the respective coefficients on the right side:
A = 1.
So, the particular solution is:
y_p = e^(2t)/(e^t + 1).
The general solution is the sum of the homogeneous solution and particular solution:
y(t) = y_h(t) + y_p(t).
That's how you can solve the given differential equation.