Hank's question at Yahoo Answers regarding a first order linear IVP

[MarkFL]

Here is the question:

Could Someone please take a look at this Calc 2 problem?!? Thank you so much!?


Solve 3 x y' - 6 y = x^{-4}, y(1) = -8.

(a) Identify the integrating factor, alpha(x).

I know in a) alpha(x) = e^(-2ln(abs(x)))

But I am not sure how to do parts (b) and (c).

(b) Find the general solution.

y(x) = ?

(c) Solve the initial value problem y(1) = -8

y(x) = ?

Thank you so much for your help!

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Hank,

First, let's write the ODE in standard linear form:

\(\displaystyle \frac{dy}{dx}+\frac{-2}{x}y=\frac{1}{3x^5}\)

a) We find the integrating factor is:

\(\displaystyle \alpha(x)=e^{-2\int\frac{dx}{x}}=\frac{1}{x^2}\)

b) Multiplying the ODE in linear form by this integrating factor, we obtain:

\(\displaystyle \frac{1}{x^2}\frac{dy}{dx}+\frac{-2}{x^3}y=\frac{1}{3x^7}\)

Observing that the left side is now the differentiation of the product of the integrating factor and $y$, we may write:

\(\displaystyle \frac{d}{dx}\left(\frac{y}{x^2} \right)=\frac{1}{3x^7}\)

Integrating both sides with respect to $x$, we get:

\(\displaystyle \int\,d\left(\frac{y}{x^2} \right)=\frac{1}{3}\int x^{-7}\,dx\)

\(\displaystyle \frac{y}{x^2}=-\frac{1}{18x^6}+C\)

Solving for $y$, we obtain the general solution:

\(\displaystyle y(x)=-\frac{1}{18x^4}+Cx^2\)

c) Using the initial values, we may write:

\(\displaystyle y(1)=-\frac{1}{18}+C=-8\implies C=-\frac{143}{18}\)

And so the solution satisfying the given IVP is:

\(\displaystyle y(x)=-\frac{1}{18x^4}-\frac{143}{18}x^2=-\frac{143x^6+1}{18x^4}\)