- #1

jegues

- 1,097

- 3

## Homework Statement

[tex] xy^{'} - 2y = x^{5} [/tex]

## Homework Equations

[tex]e^{\int P(x)dx}[/tex]

## The Attempt at a Solution

Rearranging the into the form,

[tex]y^{'} - P(x)y = Q(x)[/tex]

So,

[tex]y^{'} - \frac{2y}{x} = x^{4}[/tex]

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

Since [tex]e^{-2ln|x|} = \frac{1}{x^{2}}[/tex]

[tex]\frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}} = x^{2} [/tex]

Integrating both sides,

[tex]\int\left( \frac{y^{'}}{x^{2}} - \frac{2y}{x^{3}}\right)dx =\int x^{2}dx [/tex]

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

So,

[tex]\frac{1}{x^{2}}(y) = \int x^{2}dx[/tex]

[tex]\frac{1}{x^{2}}(y) = \frac{1}{3} x^{3} + C[/tex]

Finally,

[tex]y = \frac{x^{5}}{3} + Cx^{2}[/tex]

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

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