# Thread: Linear and separable equations

1. ## Linear and separable equations

Can someone help me with these, i've been looking through my University maths book but couldn't figure out how to do any of these. Any help would be great:

First one

Solve:

Dy/Dx= e^y .Sin^2 X/SecX

X^2 Dy/Dx +2xy=Cos^2X

2 D^2y/Dx^2 -5 dy/dx +2y=0

2. Hello, Mathsnewbie!

You should have been taught to recognize and solve different forms.

$\frac{dy}{dx}\:=\:\frac{e^y\sin^2\!x}{\sec x}$
Separate variables: . $e^{-y}dy \:=\:\sin^2\!x\cos x\,dx$

Integrate: . $-e^{-y} \:=\:\tfrac{1}{3}\sin^3\!x + C$ . . . etc.

$x^2\,\frac{dy}{dx} +2xy\:=\:\cos^2\!x$
This requires an Integrating Factor . . .

Divide by $x^2\!:\;\;\frac{dy}{dx} + \frac{2}{x}\,y \:=\:\frac{\cos^2\!x}{x^2}$

. . Then: . $I \;=\;e^{\int\frac{2}{x}dx} \;=\;e^{2\ln x} \;=\;e^{\ln(x^2)} \;=\;x^2$

Multiply by $I\!:\;\;x^2\,\frac{dy}{dx} + 2xy \:=\:\cos^2\!x$

And we have: . $\frac{d}{dx}\left(x^2y\right) \:=\:\cos^2\!x$

Integrate: . $x^2y \:=\:\int\cos^2\!x\,dx$ . . . etc.

$2\,\frac{d^2y}{dx^2} -5\,\frac{dy}{dx} +2y\:=\:0$

The characteristic equation is: . $2m^2 - 5m + 2 \:=\:0$

. . which factors: . $(m-2)(2m-1) \:=\:0$

. . and has roots: . $m \:=\:2,\tfrac{1}{2}$

Therefore: . $y \:=\:C_1e^{2x} + C_2e^{\frac{1}{2}x}$