# First Order Differential Equation

## Homework Statement

$$xy^{'} - 2y = x^{5}$$

## Homework Equations

$$e^{\int P(x)dx}$$

## The Attempt at a Solution

Rearranging the into the form,

$$y^{'} - P(x)y = Q(x)$$

So,

$$y^{'} - \frac{2y}{x} = x^{4}$$

Multiplying both sides by $$e^{\int P(x)dx}$$ or $$e^{-2\int \frac{dx}{x}}$$,

Since $$e^{-2ln|x|} = \frac{1}{x^{2}}$$

$$\frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}} = x^{2}$$

Integrating both sides,

$$\int\left( \frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}}\right)dx =\int x^{2}dx$$

Integrating the LHS should result in my integrating factor times y, $$e^{-2ln|x|}(y)$$ or $$\frac{1}{x^{2}}(y)$$

So,

$$\frac{1}{x^{2}}(y) = \int x^{2}dx$$

$$\frac{1}{x^{2}}(y) = \frac{1}{3} x^{3} + C$$

Finally,

$$y = \frac{x^{5}}{3} + Cx^{2}$$

Is this correct? (I don't have any type of solutions to check, so I thought I'd post)

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Why you did is correct.

This is a linear non-homogeneous ODE with non-constant coefficients, you solved it correctly with an integration factor.

Good job ! :)

P.S.: for future reference, you can check if your answer is correct by replacing y(x) and y'(x) with the answer and its derivative ;)
verify it. it should end up as 0=0, basically..

P.S.: for future reference, you can check if your answer is correct by replacing y(x) and y'(x) with the answer and its derivative ;)
verify it. it should end up as 0=0, basically..

I'll give verifying it a shot!

$$y = \frac{x^{5}}{3} + Cx^{2}$$

$$y^{'} = \frac{5x^{4}}{3} + 2Cx$$

Plugging into the original equation,

$$\frac{5x^{5}}{3} + 2Cx^{2} - \frac{2x^{5}}{3} - 2Cx^{2} = x^{5}$$

$$x^{5} = x^{5}$$

Thanks!

Char. Limit
Gold Member
$$\int\left( \frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}}\right)dx =\int x^{2}dx$$

Integrating the LHS should result in my integrating factor times y, $$e^{-2ln|x|}(y)$$ or $$\frac{1}{x^{2}}(y)$$

So,

$$\frac{1}{x^{2}}(y) = \int x^{2}dx$$

$$\frac{1}{x^{2}}(y) = \frac{1}{3} x^{3} + C$$

Just a little note on this part here (which is correct, I'm not saying it isn't)...

Setting, say, u = 1/x^2, we can see that u' = -2/x^3...

and so the LHS would equal u y' + y u'

which obviously equals (u y)' or (y/x^2)'

Just saying, although you probably already knew that. Just clarifying for anyone else who might read.