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)
(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!
since e^lnu = u
and (e^a)^b = e^(ab)
e^(5ln(x+4)) = (e^ln(x+4))^5 = (x+4)^5
That help?
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...
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
...
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.
To solve part (2) of the problem, we'll use the integrating factor μ(x) that you correctly found in part (1): μ(x) = e^(5ln(x+4)).
Now, we'll multiply both sides of the original equation by μ(x) to eliminate the derivative term:
(x+4)^2y' + 5(x+4)y = 16
μ(x) = e^(5ln(x+4))
Multiplying the equation by μ(x):
(x+4)^2y'e^(5ln(x+4)) + 5(x+4)ye^(5ln(x+4)) = 16e^(5ln(x+4))
Using the property of exponents, we can simplify:
(x+4)^2y(e^(5ln(x+4)))' + 5(x+4)y(e^(5ln(x+4))) = 16e^(5ln(x+4))
Here, we can apply the chain rule for the first term:
(x+4)^2y(e^(5ln(x+4)))' + 5(x+4)ye^(5ln(x+4)) = 16e^(5ln(x+4))
(x+4)^2y * 5/(x+4) + 5(x+4)y *e^(5ln(x+4)) = 16e^(5ln(x+4))
Simplifying further:
5(x+4)y + 5(x+4)y *e^(5ln(x+4)) = 16e^(5ln(x+4))
Combining like terms:
5(x+4)y(1 + e^(5ln(x+4))) = 16e^(5ln(x+4))
Now, we'll divide both sides by (1 + e^(5ln(x+4))):
5(x+4)y = 16e^(5ln(x+4))/(1 + e^(5ln(x+4)))
Simplifying the right-hand side:
5(x+4)y = 16
Finally, we can solve for y:
(x+4)^5y = 16/5
So, the explicit general solution with the arbitrary constant C is:
(x+4)^5y = 16/5 + C
Moving on to part (3) of the problem, where we need to solve the initial value problem y(0) = 4:
Substitute x = 0 and y = 4 into the general solution:
(0+4)^5*4 = 16/5 + C
Simplifying:
4^5 * 4 = 16/5 + C
1024 = 16/5 + C
To solve for C, subtract 16/5 from both sides of the equation:
C = 1024 - 16/5
C = 5114/5
Therefore, the solution to the initial value problem is:
(x+4)^5y = 16/5 + 5114/5
Simplifying:
(x+4)^5y = (5130/5)
Thus, the solution to the initial value problem y(0) = 4 is:
y = (5130/[(x+4)^5])