find-a-particular-solution-to-the-equation-y-y-x-sin-y-x-with-original-condition-y-1-pi-2-

Question Number 96192 by 1549442205 last updated on 30/May/20

find a particular solution to the equation

y^{'} = \frac{y}{x} + \sin \frac{y}{x} with original condition

y (1) = \frac{π}{2}

Commented by john santu last updated on 30/May/20

set v = \frac{y}{x} \Rightarrow \frac{dy}{dx} = v + x \frac{d v}{d x}

v + x \frac{d v}{d x} = v + \sin v

x \frac{d v}{d x} = \sin v \Rightarrow \int \frac{d v}{\sin v} = \int \frac{d x}{x}

\int \csc v d v = \ln (x) + c

\ln (\csc v - \cot v) = \ln C x

c s c v - \cot v = C x

\frac{1 - \cos (\frac{y}{x})}{\sin (\frac{y}{x})} = C x

1 - \cos (\frac{y}{x}) = C x \sin (\frac{y}{x})

Commented by 1549442205 last updated on 11/Jun/20

This is an other way

Put \frac{y}{x} = t \Rightarrow y = tx \Rightarrow dy = xdt + tdx . Hence,

xdt + tdx = (t + sint) dx \Rightarrow xdt = sintdx

\Rightarrow \frac{dt}{sint} = \frac{dx}{x} . Integrate two sides we get

\ln ∣ \tan (\frac{t}{2}) ∣= \ln ∣ x ∣ + lnC . From that ∣

\frac{t}{2} = \arctan (Cx) \Rightarrow y = 2 x . \arctan (Cx)

Using the original condition we get

\frac{π}{2} = 2 arctanC \Rightarrow C = 1 . Thus, the

particular solution has form :

\boldsymbol y = 2 \boldsymbol x . \boldsymbol arctanx