Question
Solve the following differential equations
a). sin -1(dy/dx) = x + y
b). (1 + y2)dx = (tan-1 y - x)dy
c). (D - 1)2(D2 + 1)2 y = sin2
d).
Answer :
Word Count : 777
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Skip in 10
a) Given: $$ \sin^{-1}\left(\frac{dy}{dx}\right) = x + y $$ Let $p = \frac{dy}{dx}$. Then $$ \sin^{-1}p = x + y \implies p = \sin(x + y) $$ So we have $$ \frac{dy}{dx} = \sin(x + y) $$ Set $v = x + y \implies \frac{dv}{dx} = 1 + \frac{dy}{dx} = 1 + \sin v$ $$ \frac{dv}{dx} = 1 + \sin v \implies \frac{dv}{1 + \sin v} = dx $$ Use identity: $1 + \sin v = (\sin v/2 + \cos v/2)^2$ or more simply, $\frac{1}{1+\sin v} = \csc v/ ( \csc v + \cot v )$. Better: multiply numerator and denominator by $1 - \sin v$: $$ \frac{dv}{1 + \sin v} = \frac{1 - \sin v}{\cos^2 v} dv = \frac{\cos^2 v}{\cos^2 v} dv? \] Simpler: standard integral: \(\int \frac{dv}{1 + \sin v} = \int \frac{1 - \sin v}{\cos^2 v} dv = \int \sec^2 v (1 - \sin v) dv\) \[ \int \frac{1 - \sin v}{\cos^2 v} dv = \int \sec^2 v dv - \int \frac{\sin v}{\cos^2 v} dv = \tan v
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a). sin -1(dy/dx) = x + y
b). (1 + y2)dx = (tan-1 y - x)dy
c). (D - 1)2(D2 + 1)2 y = sin2
d).