Question

y'' -4y' +3y =e^(5t) using variation of parameters

Answer 1

the solution for this will be in form:
u(t)=u_c(t)+U_p(t)
where u_c(t) is a complemntary solution and U_p(t) will be particular solution...
ok so first find complementary solution u_c(t) or solution for:
y''-4y'+3y=0
now the roots of the characteristic equation of this will be 1 and 3 and so the particular solution is just
u_c(t)=c_1e^1t+c_2e^3t
so our y_1=e^t and y_2=e^3t
now we from variation of parameters that
\[U_p(t)=-y_1\int\limits y_2g(t)/W(y_1,y_2)dt+y2 \int\limits y_1g(t)/W(y_1,y_2)dt\]
where g(t) is the terms in the right had side of the equation or in this case e^5t and W(y_1,y_2) is the wronskian of y_1 and y_2. so the wronskian will be:
W(y_1,y_2)=y_1y_2'-y_2y_1'=e^t*3e^3t-e^3t*e^t=3e^4t-e^4t=2e^4t
so our U_p(t) will be:
\[U_p(t)=-e^t \int\limits e^{3t}e^{5t}/2e^{4t}dt+e^{3t} \int\limits e^te^{5t}/2e^{4t}dt\]
\[U_p(t)=-e^t/2 \int\limits\limits e^{4t}dt+e^{3t}/2 \int\limits\limits e^{2t}dt\]
\[U_p(t)=-e^{5t}/8+e^{5t}/4=e^{5t}/8\]
and so our general solution will be:
\[u(t)=c_1e^t+c_2e^{3t}+e^{5t}/8\]

Answer 2

Thank you very much and this helps alot! I understand every step as well! i just worked it out and applies this to all my other problems I have and understand them all. Thanks!