Linear First Order ODE/x y' + y = x^2 cosine x/Proof 1
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Theorem
- $x \, \dfrac {\d y} {\d x} + y = x^2 \cos x$
has the general solution:
- $y = 2 \cos x + x \sin x - \dfrac 2 x \sin x + \dfrac C x$
Proof
Rearranging:
- $\dfrac {\d y} {\d x} + \dfrac y x = x \cos x$
This is in the form:
- $\dfrac {\d y} {\d x} + \map P x y = \map Q x$
where:
- $\map P x = \dfrac 1 x$
- $\map Q x = x \cos x$
Thus:
\(\ds \int \map P x \rd x\) | \(=\) | \(\ds \int \dfrac 1 x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \ln x\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds e^{\int P \rd x}\) | \(=\) | \(\ds x\) |
Thus from Solution by Integrating Factor:
- $\map {\dfrac {\d} {\d x} } {x y} = x^2 \cos x$
\(\ds \map {\dfrac {\d} {\d x} } {x y}\) | \(=\) | \(\ds x^2 \cos x\) | ||||||||||||
\(\ds x y\) | \(=\) | \(\ds \int x^2 \cos x \rd x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 2 x \cos x + \paren {x^2 - 2} \sin x + C\) | Primitive of $x^2 \cos a x$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds 2 \cos x + x \sin x - \frac 2 x \sin x + \frac C x\) |
$\blacksquare$