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.
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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.
For clarity 3rd question is rewritten as
SOLVE
(x*y^2)dx - e^(1/(x^3))dx - (x^2)*y dy = 0
Hello, ceode!
Quote:
$\displaystyle \text{(1) Solve: }\;\dfrac{dy}{dx} + y\cdot\cot x \:=\:5e^{\cos x},\quad y\left(\tfrac{5}{4}\right) \,=\,-4$
$\displaystyle \text{Integrating factor: }\;I \:=\:e^{\int\cot x\,dx} \;=\;e^{\ln(\sin x)} \;=\;\sin x$
$\displaystyle \text{Multiply by }I\!:\;\;\sin x\cdot\frac{dy}{dx} + y\cdot\cot x\cdot\sin x \;=\;5\sin x\cdot e^{\cos x}$
$\displaystyle \text{We have: }\;\dfrac{d}{dx}\left(\sin x\cdot y\right) \;=\;5\sin x\cdot e^{\cos x}$
$\displaystyle \displaystyle \text{Integrate: }\;\int\frac{d}{dx}(\sin x\cdot y) \;=\;5\int\sin x\cdot e^{\cos x}dx$
. . . . . . . . . . . . . .$\displaystyle y\cdot \sin x \;=\;-5e^{\cos x} + C$
. . . . . . . . . . . . . . . . . . $\displaystyle y \;=\;\dfrac{C - 5e^{\cos x}}{\sin x} $
$\displaystyle 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} $
$\displaystyle \text{Therefore: }\;y \;=\;\dfrac{5e^{\cos\frac{5}{4}} - 4\sin\frac{5}{4} - 5e^{\cos x}}{\sin x} $
Thank you sir , can you help me with the other 2 questions
