Linear and separable equations

• Mar 12th 2009, 05:41 AM
Mathsnewbie
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
• Mar 12th 2009, 06:17 AM
Soroban
Hello, Mathsnewbie!

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

Quote:

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

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

Quote:

$x^2\,\frac{dy}{dx} +2xy\:=\:\cos^2\!x$
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.

Quote:

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

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