The solution of d y d x + x y = x 2 is Options: A . &nbsp 1 y = c x − x l o g x B . &nbsp 1 x = c y − y l o g y C . &nbsp 1 x = c x − x l o g y D . &nbsp 1 y = c x − y l o g x

Answer: Option B : B d x d y + x y = x 2 ⇒ x − 2 d x d y + x − 1 y = 1 Put x − 1 = t . T h e n − x − 2 d x d y = d t d y ⇒ x − 2 d x d y = − d t d y ∴ − d t d y + t y = 1 ⇒ d t d y − 1 y t = − 1 It is linear in t. Here P = − 1 y . Q = − 1 I . F = e ∫ ( 1 y ) d y = e − l o g y = 1 y ∴ Solution is t ( 1 y ) = ∫ ( − 1 ) 1 y d y = − l o g y + c ⇒ t = − y l o g y + c y ⇒ x − 1 c y − y l o g y .

The solution of dydx+xy=x2 is

Discussion Forum : Differential Equations

Question -

The solution of $\frac{d y}{d x} + \frac{x}{y} = x^{2}$ is

Options:

A .

\frac{1}{y} = c x - x l o g x

B .

\frac{1}{x} = c y - y l o g y

C .

\frac{1}{x} = c x - x l o g y

D .

\frac{1}{y} = c x - y l o g x

Answer: Option B
:
B

\frac{d x}{d y} + \frac{x}{y} = x^{2} \Rightarrow x^{- 2} \frac{d x}{d y} + \frac{x^{- 1}}{y} = 1

Put

x^{- 1} = t . T h e n - x^{- 2} \frac{d x}{d y} = \frac{d t}{d y} \Rightarrow x^{- 2} \frac{d x}{d y} = - \frac{d t}{d y} ∴ - \frac{d t}{d y} + \frac{t}{y} = 1 \Rightarrow \frac{d t}{d y} - \frac{1}{y} t = - 1

It is linear in t. Here

P = - \frac{1}{y} . Q = - 1

I . F = e^{\int (\frac{1}{y}) d y} = e^{- l o g y} = \frac{1}{y}

∴

Solution is

t (\frac{1}{y}) = \int (- 1) \frac{1}{y} d y = - l o g y + c \Rightarrow t = - y l o g y + c y \Rightarrow x^{- 1} c y - y l o g y .

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