# Ordinary differential equations

• Oct 8th 2010, 10:52 AM
ceode
Ordinary differential equations
1 ) SOLVE
dy/dx + y*cot(x) =5*e^cos(x) if y=-4 and x=5/4

2)
solve

dy/dx + (y logy)/x = y(logy)^2/(x^2)

3)SOLVE
(x*y^2)dx - e^(1/(x^3))dx - x^2*y dy = 0

Thank you very much.
• Oct 8th 2010, 10:56 AM
ceode
For clarity 3rd question is rewritten as

SOLVE
(x*y^2)dx - e^(1/(x^3))dx - (x^2)*y dy = 0
• Oct 8th 2010, 12:44 PM
Soroban
Hello, ceode!

Quote:

$\text{(1) Solve: }\;\dfrac{dy}{dx} + y\cdot\cot x \:=\:5e^{\cos x},\quad y\left(\tfrac{5}{4}\right) \,=\,-4$

$\text{Integrating factor: }\;I \:=\:e^{\int\cot x\,dx} \;=\;e^{\ln(\sin x)} \;=\;\sin x$

$\text{Multiply by }I\!:\;\;\sin x\cdot\frac{dy}{dx} + y\cdot\cot x\cdot\sin x \;=\;5\sin x\cdot e^{\cos x}$

$\text{We have: }\;\dfrac{d}{dx}\left(\sin x\cdot y\right) \;=\;5\sin x\cdot e^{\cos x}$

$\displaystyle \text{Integrate: }\;\int\frac{d}{dx}(\sin x\cdot y) \;=\;5\int\sin x\cdot e^{\cos x}dx$

. . . . . . . . . . . . . . $y\cdot \sin x \;=\;-5e^{\cos x} + C$

. . . . . . . . . . . . . . . . . . $y \;=\;\dfrac{C - 5e^{\cos x}}{\sin x}$

$y\left(\frac{5}{4}\right) \,=\,-4\!:\;\; \dfrac{C - 5e^{\cos\frac{5}{4}}}{\sin\frac{5}{4}}} \;=\;-4 \quad\Rightarrow\quad C \;=\;5e^{\cos\frac{5}{4}} - 4\sin\frac{5}{4}$

$\text{Therefore: }\;y \;=\;\dfrac{5e^{\cos\frac{5}{4}} - 4\sin\frac{5}{4} - 5e^{\cos x}}{\sin x}$
• Oct 8th 2010, 07:31 PM
ceode
Thank you
Thank you sir , can you help me with the other 2 questions