Find-the-differential-equations-i-log-dy-dx-ax-by-ii-x-cos-y-dy-x-e-x-log-x-e-x-dx-

Question Number 83639 by niroj last updated on 04/Mar/20

Find the differential equations :

(i) \log (\frac{dy}{dx}) = ax + by

(ii) x \cos y dy = (x e^{x} \log x + e^{x}) dx

Answered by TANMAY PANACEA last updated on 05/Mar/20

1) l n (\frac{d y}{d x}) = a x + b y

\frac{d y}{d x} = e^{a x + b y}

\frac{d y}{e^{b y}} = e^{a x} d x

\int e^{- b y} d y = \int e^{a x} d x

\frac{e^{- b y}}{- b} = \frac{e^{a x}}{a} + c

Commented by niroj last updated on 05/Mar/20

i t s ★ g r e a t

Commented by TANMAY PANACEA last updated on 05/Mar/20

m o s t w e l c o m e

Answered by TANMAY PANACEA last updated on 05/Mar/20

2) x c o s y d y - ({x e}^{x} l n x + e^{x}) d x

\int c o s y d y - \int (e^{x} l n x + \frac{e^{x}}{x}) d x = 0

\int c o s y d y - \int \frac{d}{d x} (e^{x} l n x) d x = 0

s i n y - e^{x} l n x = c

Commented by niroj last updated on 05/Mar/20

t h a n k y o u p a n a c e a .

Commented by TANMAY PANACEA last updated on 05/Mar/20

m o s t w e l c o m e

Answered by M±th+et£s last updated on 05/Mar/20

l o g (\frac{d y}{d x}) = a x + b y \to l o g (p) = a x + b y \to y = \frac{1}{b} (l o g (p) - a x)

D i f f . w . r . t . x \to p = \frac{1}{b} (\frac{d l o g (p)}{d x} - a)

p = \frac{1}{b} (\frac{1}{p} \frac{d p}{d x} - a) \to \frac{d x}{d p} = \frac{1}{p (a + b p)} = \frac{1}{a} \frac{1}{p} - \frac{b}{a} \frac{1}{a + b p}

x = \frac{1}{a} l n ∣ p ∣ - \frac{1}{a} l n ∣ a + b p ∣ + \frac{1}{a} l n (c)

x = \frac{1}{a} l n (\frac{c p}{a + b p}) \to e^{a x} = \frac{c p}{a + b p} \to p = \frac{{a e}^{a x}}{c - {b e}^{a x}}

y = - \frac{l n (c - {b e}^{a x})}{b}

Answered by M±th+et£s last updated on 05/Mar/20

c o s (y) d y = \frac{{x e}^{x} l o g (x) + e^{x}}{x} d x

\to s i n (y) = \int (e^{x} l o g (x) + \frac{e^{x}}{x}) d x = \int e^{x} l o g (x) d x + e^{x} l o g (x) - \int e^{x} l o g (x) d x

s i n (y) = e^{x} l o g (x) + c \to y = {s i n}^{- 1} (c + e^{x} l o g (x))