Linear First Order ODE/x dy + y dx = x cosine x dx

From ProofWiki
Jump to navigation Jump to search

Theorem

The linear first order ODE:

$x \rd y + y \rd x = x \cos x \rd x$

has the general solution:

$x y = x \sin x + \cos x + C$


Proof 1

Rearranging:

$\dfrac {d y} {\d x} + \dfrac y x = \cos x$


This is a linear first order ODE in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where:

$\map P x y = \dfrac 1 x$
$\map Q x = \cos x$


Thus:

\(\ds \int \map P x \rd x\) \(=\) \(\ds \int \frac 1 x \rd x\)
\(\ds \) \(=\) \(\ds \ln x\)
\(\ds \leadsto \ \ \) \(\ds e^{\int P \rd x}\) \(=\) \(\ds e^{\ln x}\)
\(\ds \) \(=\) \(\ds x\)


Thus from Solution by Integrating Factor:

\(\ds \dfrac {\d} {\d x} \paren {x y}\) \(=\) \(\ds x \cos x\)
\(\ds \leadsto \ \ \) \(\ds x y\) \(=\) \(\ds \int x \cos x \rd x\)
\(\ds \) \(=\) \(\ds x \sin x + \cos x + C\) Primitive of $x \cos a x$

$\blacksquare$


Proof 2

\(\ds x \dfrac {\d y} {\d x} + y\) \(=\) \(\ds x \cos x\)
\(\ds \leadsto \ \ \) \(\ds x y\) \(=\) \(\ds \int x \cos x \rd x + C\) Linear First Order ODE: $x y' + y = \map f x$
\(\ds \) \(=\) \(\ds x \sin x + \cos x + C\) Primitive of $x \cos a x$

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Linear_First_Order_ODE/x_dy_%2B_y_dx_%3D_x_cosine_x_dx&oldid=438053"