How to solve this question 4 Suppose y Ser and y e nonhomoge
 How to solve this question?
Solution
a)
In the two given solutions coefficient of t^2 remains unchanged so that is the particular solution
and the two exponential components of y1 and y2 are solutions to the homogeneous equations ie
y\'\'+py\'+qy=0
a)
No this is not a solution
Coeffficient of particular solution remains unchanged. Only the coefficients of the exponetial terms ie solution to homgeneous equation can change
b) Yes. It is a particular solution
c)
General solution is
y=A e^{2t}+B e^{-5t}-6t^2
d)
First we set up the homogeneous equation
e^{2t} amd e^{-5t} are solutions so characterisitc is
(k-2)(k+5)=k^2+3k-10
Hence homogeneous ode is
y\'\'+3y\'-10=0
So the nonhomogeneous ode is
y\'\'+3y\'-10y=g(t)
-6t^2 is a solution so we substitute and then compute g(t)
-(6t^2)\'\'-3(6t^2)\'+10*6t^2=g(t)
-12-36t+60t^2=g(t)
y\'\'+3y\'-10y=60t^2-36t-12
p(t)=3, q(t)=-10,g(t)=60t^2-36t-12

