If-a-curve-y-f-x-passing-through-the-point-1-2-is-the-solution-of-differential-equation-2x-2-dy-2xy-y-2-dx-then-the-value-of-f-2-is-equal-to-

Question Number 118376 by bramlexs22 last updated on 17/Oct/20

I f a c u r v e y = f (x) p a s s i n g t h r o u g h

t h e p o i n t (1, 2) i s t h e s o l u t i o n

o f d i f f e r e n t i a l e q u a t i o n

2 x^{2} d y = (2 x y + y^{2}) d x, t h e n t h e

v a l u e o f f (2) i s e q u a l t o ?

Answered by benjo_mathlover last updated on 17/Oct/20

s o l v i n g f o r d i f f e r e n t i a l e q u a t i o n

\frac{d y}{d x} = \frac{2 x y + y^{2}}{2 x^{2}} . [s e t y = z x]

\Rightarrow x \frac{d z}{d x} + z = \frac{2 x (z x) + z^{2} x^{2}}{2 x^{2}}

\Rightarrow x \frac{d z}{d x} + z = \frac{2 z + z^{2}}{2}

\Rightarrow x \frac{d z}{d x} = \frac{1}{2} z^{2}; \frac{d z}{z^{2}} = \frac{1}{2} \frac{d x}{x}

\Rightarrow - \frac{1}{z} = \frac{1}{2} \ln (x) + c; o r

\Rightarrow \frac{x}{y} = - \frac{1}{2} \ln (x) - c; s u b s t i t u t e p o i n t (1, 2)

\Rightarrow \frac{1}{2} = - \frac{1}{2} \ln (1) - c; c = - \frac{1}{2}

t h u s \frac{x}{y} = - \frac{1}{2} \ln (x) + \frac{1}{2}

\Rightarrow \frac{2 x}{y} = 1 - \ln (x) o r y = \frac{2 x}{1 - \ln (x)}

t h e r e f o r e f (2) = \frac{4}{1 - \ln (2)}

Answered by TANMAY PANACEA last updated on 17/Oct/20

2 x^{2} d y - 2 x y d x = y^{2} d x

- 2 x (\frac{y d x - x d y}{y^{2}}) = d x

- 2 d (\frac{x}{y}) = \frac{d x}{x}

- 2 (\frac{x}{y}) = l n x + l n c

l n (x c) = \frac{- 2 x}{y}

x c = e^{- \frac{2 x}{y}}

1 \times c = e^{- \frac{2 \times 1}{2}} \to \boldsymbol c = \boldsymbol e^{- 1}

\boldsymbol {x e}^{- 1} = \boldsymbol e^{- \frac{2 \boldsymbol x}{\boldsymbol y}}

(\boldsymbol l n x) + (- 1) = \frac{- 2 \boldsymbol x}{\boldsymbol y}

\boldsymbol y = \frac{- 2 \boldsymbol x}{- 1 + \boldsymbol l n x} \to f (2) = \frac{- 4}{- 1 + \boldsymbol l n 2} = \frac{4}{1 - \boldsymbol l n 2}