Solve the differential equation $(x - y^{2}) d x + 2 x y d y = 0$ .

Question

Verified by Toppr

Answer 1

$(x - y^{2}) d x + 2 x y d y = 0 \Rightarrow 2 y \frac{d y}{d x} - \frac{y ^{2}}{x} = - 1$
Substitute $v = y^{2} \Rightarrow d v = 2 y d y$
$\frac{d v}{d x} - \frac{v}{x} = - 1$ ...(1)
Here $P = - \frac{1}{x} \Rightarrow \int P d x = - \int \frac{1}{x} d x = - lo g x = lo g \frac{1}{x}$
$∴ I . F . = e^{l o g \frac{1}{x}} = \frac{1}{x}$
Multiplying (1) by $I . F .$ we get
$\frac{1}{x} \frac{d v}{d x} - \frac{v}{x ^{2}} = - \frac{1}{x}$
Integrating both sides w.r.t $x$ we get
$\frac{v}{x} = - \int \frac{1}{x} d x + a = - lo g x + a \Rightarrow y^{2} = x lo g a x$

Answer 2

(x - y^{2}) d x + 2 x y d y = 0 \Rightarrow 2 y \frac{d y}{d x} - \frac{y ^{2}}{x} = - 1

Substitute

v = y^{2} \Rightarrow d v = 2 y d y

\frac{d v}{d x} - \frac{v}{x} = - 1

...(1)
Here

P = - \frac{1}{x} \Rightarrow \int P d x = - \int \frac{1}{x} d x = - lo g x = lo g \frac{1}{x}

∴ I . F . = e^{l o g \frac{1}{x}} = \frac{1}{x}

Multiplying (1) by

I . F .

we get

\frac{1}{x} \frac{d v}{d x} - \frac{v}{x ^{2}} = - \frac{1}{x}

Integrating both sides w.r.t

x

we get

\frac{v}{x} = - \int \frac{1}{x} d x + a = - lo g x + a \Rightarrow y^{2} = x lo g a x