# SOLVED Integration by Parts Question

• Monocles
In summary, the problem given is to find the solution of a first order differential equation involving integration by parts. The attempt at a solution involves manipulating the equation and using the integration by parts method, but the person gets stuck at a certain point. However, they are able to solve the problem with the help of a commenter pointing out a simple error in their calculations.
Monocles
[SOLVED] integration by parts question

## 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
$$x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)$$
with
$$y(\pi) = 0$$

## Homework Equations

Format of first order differential equations:
$$y' + P(x) y = q(x)$$

$$H(x) = \int P(x) dx$$

## The Attempt at a Solution

$$x^3 y'+ax^2 y = 2x^{-a+4} \sin(x)$$

$$y' + \frac{a y}{x} = \frac{2 x^{-a + 4} \sin(x)}{x^3}$$

$$y' + \frac{a y}{x} = 2 x^{-a + 1} \sin(x)$$

$$H(x) = \int \frac{a}{x} dx$$

$$H(x) = a \ln{x}$$

Take $$e^{H(x)} = x^a$$ and multiply it to both sides

$$x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)$$

$$(x^a y)' = 2 x^{-a^2 + a} \sin(x)$$

$$x^a y = \int 2 x^{-a^2 + a} \sin(x) dx$$

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.

Last edited:
$$x^a y' + \frac{a y}{x} x^a = x^a 2 x^{-a + 1} \sin(x)$$

$$(x^\alpha\,y)'=2\,x\,\sin x$$

Ohh... Well that was easy. I just forgot how to multiply! Thanks.

## 1. How do you determine which function to integrate by parts?

The general rule is to choose the function that will become simpler when differentiated and the other function should become simpler when integrated.

## 2. What is the formula for integration by parts?

The formula is ∫u dv = uv - ∫v du, where u and v are the two functions being integrated and dv and du are their respective differentials.

## 3. What is the purpose of using integration by parts?

Integration by parts is used to simplify the integration of a product of two functions that cannot be easily integrated using other methods.

## 4. Can integration by parts be used to solve definite integrals?

Yes, integration by parts can be used to solve definite integrals. In this case, the formula is modified to ∫a^b u dv = uv|a^b - ∫a^b v du, where a and b are the upper and lower limits of integration.

## 5. Are there any guidelines for choosing which function to differentiate and which to integrate?

There are no strict guidelines, but it is generally helpful to choose the function that contains an algebraic term as the one to differentiate and the function that contains a trigonometric or exponential term as the one to integrate.

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