Solve-the-differential-equation-y-cosh-x-y-

Question Number 25572 by hoangnampham13 last updated on 11/Dec/17

S o l v e t h e d i f f e r e n t i a l e q u a t i o n :

y^{'} = c o s h (x + y)

Answered by prakash jain last updated on 11/Dec/17

y^{'} = c o s h (x + y) (A)

u (x) = y (x) + x

u^{'} (x) = y^{'} (x) + 1

s u b t i t u t i n g y^{'} f r o m A

u^{'} (x) = \cosh (x + y) + 1

u^{'} (x) = \cosh (u) + 1

\frac{d u}{d x} = \cosh u + 1

\frac{d u}{\cosh u + 1} = d x

\int \frac{d u}{\cosh u + 1} = \int d x

v = \tanh \frac{u}{2}

d v = \frac{1}{2} {sech}^{2} \frac{u}{2} d u = \frac{d u}{{2 \cosh}^{2} \frac{u}{2}}

\cosh u + 1 = {2 \cosh}^{2} \frac{u}{2}

\int \frac{d u}{\cosh u + 1} = \int d v = v + c = \tanh \frac{u}{2} + c

\tanh \frac{u}{2} + c = x

\tanh \frac{u}{2} = x - c

\frac{u}{2} = \tanh^{- 1} (x - c)

u = {2 \tanh}^{- 1} (x - c)

y (x) + x = {2 \tanh}^{- 1} (x - c)

y (x) = {2 \tanh}^{- 1} (x - c) - x