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[SOLVED] integration by parts question
I think my first problem is at integration by parts, but let me know if you see a different error in my work.
Edit: I shoud mention that this is Calculus 2 and we just learned first order differential equations today. So I don't have any fancy methods available.
Find the solution of
[tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]
with
[tex]y(\pi) = 0[/tex]
Format of first order differential equations:
[tex]y' + P(x) y = q(x)[/tex]
[tex]H(x) = \int P(x) dx[/tex]
[tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]
[tex]y' + \frac{a y}{x} = \frac{2 x^{-a + 4} \sin(x)}{x^3}[/tex]
[tex]y' + \frac{a y}{x} = 2 x^{-a + 1} \sin(x)[/tex]
[tex]H(x) = \int \frac{a}{x} dx[/tex]
[tex]H(x) = a \ln{x}[/tex]
Take [tex]e^{H(x)} = x^a[/tex] and multiply it to both sides
[tex]x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)[/tex]
[tex](x^a y)' = 2 x^{-a^2 + a} \sin(x)[/tex]
[tex] x^a y = \int 2 x^{-a^2 + a} \sin(x) dx[/tex]
This is where I am stuck. The only way I can think of to solve that integral is integration by parts, but the only way I can see that working is by repeating it enough times for the part of the integral before sin(x) to disappear. So I think I went wrong somewhere? But I don't see where.
Homework Statement
I think my first problem is at integration by parts, but let me know if you see a different error in my work.
Edit: I shoud mention that this is Calculus 2 and we just learned first order differential equations today. So I don't have any fancy methods available.
Find the solution of
[tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]
with
[tex]y(\pi) = 0[/tex]
Homework Equations
Format of first order differential equations:
[tex]y' + P(x) y = q(x)[/tex]
[tex]H(x) = \int P(x) dx[/tex]
The Attempt at a Solution
[tex]x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)[/tex]
[tex]y' + \frac{a y}{x} = \frac{2 x^{-a + 4} \sin(x)}{x^3}[/tex]
[tex]y' + \frac{a y}{x} = 2 x^{-a + 1} \sin(x)[/tex]
[tex]H(x) = \int \frac{a}{x} dx[/tex]
[tex]H(x) = a \ln{x}[/tex]
Take [tex]e^{H(x)} = x^a[/tex] and multiply it to both sides
[tex]x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)[/tex]
[tex](x^a y)' = 2 x^{-a^2 + a} \sin(x)[/tex]
[tex] x^a y = \int 2 x^{-a^2 + a} \sin(x) dx[/tex]
This is where I am stuck. The only way I can think of to solve that integral is integration by parts, but the only way I can see that working is by repeating it enough times for the part of the integral before sin(x) to disappear. So I think I went wrong somewhere? But I don't see where.
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