Respuesta :
The question is not well-typed, but I understand that:
A first order linear equation in the form
y′ + p(x)y = f(x) ...........................(1)
can be solved by finding an integrating factor
μ(x) = e^(∫p(x)dx) .......................(2)
(1) Given the equation
(x + 2)²y′ + 5(x + 2)y = 16 ..........(3)
find μ(x).
SOLUTION:
First, we need the equation (3) to be in the form of (1). To do this, we will divide the equation by (x + 2)². So we have
y′ + 5/(x + 2)y = 16/(x + 2)²
This is now in the form of (1), with
p(x) = 5/(x + 2), and f(x) = 16/(x + 2)² .
The integrating factor can now be obtained using (2).
μ(x) = e^{∫[5/(x + 2)]dx}
= e^[5ln(x + 2)]
μ(x) = (x + 2)^5
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(2) Then an explicit general solution with arbitrary constant C can be written in the form ((x + 2)^5)y= f(x) + C.
SOLUTION
To obtain an explicit general solution of (3), first, we multiply (3) by the integrating factor. Doing that, we have:
((x + 2)^5)y′ + 5((x + 2)^4)y = 16(x + 2)³
This can be expressed as
d(y(x + 2)^5) = 16(x + 2)³
Integrating both sides, we have
y(x + 2)^5 = 4(x + 2)^4 + C........(4)
Which is the form we are required to have.
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(3) Then solve the initial value problem with y(0) = 4
SOLUTION
Given the condition y(0) = 4, we put y = 4, and x = 0 in the general solution (4) obtained to obtain the value of the arbitrary constant C.
Doing that, we have
4(0 + 2)^5 = 4(0 + 2)^4 + C
C = 4(2^5) - 4(2^4)
= 128 - 64 = 64
Finally, we insert C = 64 in (4) to have the required particular solution. Doing that, we have
y(x + 2)^5 = 4(x + 2)^4 + 64
And that is it.
