A first order linear equation in the form y′+p(x)y=f(x) can be solved by finding an integrating factor μ(x)=exp(∫p(x)dx)

Question

(1) Given the equation (x+4)^2y′+5(x+4)y=16 find μ(x)= I found it to be e^(5ln(x+4))which is correct**

(2) Then an explicit general solution with arbitrary constant C can be written in the form:

(x+4)^5y=__________ +C.

(3) Then solve the initial value problem with y(0)=4

y=________

I need help with parts two and three. I cant figure out how to get the equation into the form for question two. Without the answer for two the answer for question three is impossible. Please help!

Answer 1

since e^lnu = u
and (e^a)^b = e^(ab)

e^(5ln(x+4)) = (e^ln(x+4))^5 = (x+4)^5

That help?

Answer 2

well when I set it up i get Ce^(-5ln(x+4))*integral(e^(5ln(x+4)*(16/(x+4)^2) which i think brings me to Ce^(-5ln(x+4))*(4(x+4)^4) which would simplify to: Ce^(-5ln(x+4))+(4/(x+4)) am I doing the process right because this is where I am getting stuck. I understand the relationship you posted but not how it helps me...

Answer 3

we have
(x+4)^2y′+5(x+4)y=16
y' + 5/(x+4) y = 16/(x+4)^2

The integrating factor is

e^(∫5/(x+4) dx) = (x+4)^5

So, multiplying through by that we have

(x+4)^5 y' + 5(x+4)^4 y = 16(x+4)^3

d((x+4)^5 y = 16(x+4)^3
x+4)^5 y = 4(x+4)^3 + C
...

Answer 4

sorry - had to run off - also had a typo above

d((x+4)^5 y) = 16(x+4)^3
(x+4)^5 y = c + 4(x+4)^4
y = c/(x+4)^5 + 4/(x+4)

Now you can solve the initial value problem.