Linear First Order ODE/(1 + x^2) y' + 2 x y = 4 x^3

Theorem

The linear first order ODE:

$\paren {1 + x^2} \dfrac {\d y} {\d x} + 2 x y = 4 x^3$

has the general solution:

$y = \dfrac {x^4} {1 + x^2} + \dfrac C {1 + x^2}$

Proof

It is noticed that:

$\dfrac {\d} {\d x} \paren {1 + x^2} = 2 x$

and so $(1)$ can be rendered in the form:

\(\ds \map \d {\paren {1 + x^2} y}\)

\(=\)

\(\ds 4 x^3\)

\(\ds \leadsto \ \ \)

\(\ds \paren {1 + x^2} y\)

\(=\)

\(\ds \int 4 x^3 \rd x\)

\(\ds \)

\(=\)

\(\ds x^4 + C\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds \frac {x^4} {1 + x^2} + \frac C {1 + x^2}\)

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $21$