Method of undertimed coefficents video
Method of Undetermined coefficents is a method to solve non-homogenous differential equations.
General solution is in the form below
y=yh+yp
where the second order non homogenous ODE has the form
a(x)y′′+b(x)y′+c(x)y=g(x)
- yh is the solution to the homogenous ODE a(x)y′′+b(x)y′+c(x)y=0
- yp, the particular solution, is a function that satisfies the non-homogenous equation. This can be determined by using the method of undetermined coefficents.
Problem 5: Find the General solution the non homogenous ODE y′′−8y′+16y=16e6t
Solution 5:
Solve for yc first.
m2−8m+16=(m−4)(m−4)⟶m=4,4
So, yc=c1e4t+c2e4t
Use method of undetermined coefficents
yp=Ae6t⟶36Ae6t−48Ae6t+16Ae4t=Ae6t(36−48+16)=Ae6t(36−48+16)
This simplifies to
4Ae6t=16e6t→A=4
So, yp=4e6t
General solution is y=c1e4t+c2e4t+4e6t
Problem 6: Find intial values and solve the DE: y′′+25y=e−t, y(0)=y0, y′(0)=y0′. Suppose we know that y(t)→0 as t→∞.
Solution 6:
Get general solution first y=yh+yp. For yh, m=±5i since characteristic equation is m2+25=0. So, yh is c1cos(5t)+c2sin(5t).
Solve for yp next. yp=Ae−t. Plug in to the equation: Ae−t+25Ae−t=e−t. Simplifies to 26Ae−t=e−t, making A=261
General solution then is y=c1cos(5t)+c2sin(5t)+261e−t.
Find initial conditions. First, we do y(0). y(0)=c1+1/26, where c1 and c2 have to be 0 here in order for limx→∞y=0, and so y(0)=1/26. We get that limit because we know that y(t)→0 as t→∞. That is our first initial condition. Now, our second initial condition y′(0)=−c1sin(5∗0)+c2cos(5∗0)+1/26=c2−1/26 and since c2 is also 0, y′(0)=−1/26. Our initial conditions are y(0)=1/26, and y′(0)=−1/26.
Now, solve for y given the initial conditions. y=261e−t from plugging in both the initial conditions and solving like a normal IVP.
Answers: y(0)=1/26, y’(0)=−1/26, and y=261e−t