(a) Solve the following differential equation, given that $y=1$ when $x=1$: $$\left(x^{2}-y^{2}\right)dx+2xy\,dy=0$$ (b) Solve the following differential equation, given that $y=1$ when $x=1$: $$\dfrac{dy}{dx}=\dfrac{x(2y-x)}{x(2y+x)}$$

(a) $x^{2}+y^{2}=2x$; (b) $\log\!\left(\dfrac{2y^{2}-xy+x^{2}}{x^{2}}\right)+\dfrac{6}{\sqrt{7}}\tan^{-1}\!\left(\dfrac{4y-x}{\sqrt{7}\,x}\right)+2\log x=\log 2+\dfrac{6}{\sqrt{7}}\tan^{-1}\dfrac{3}{\sqrt{7}}$

Differential Equations — CBSE Class 12 · 2008 Board Question with Solution

(a) Solve the following differential equation, given that

y = 1

when

x = 1

(x^{2} - y^{2}) d x + 2 x y d y = 0

(b) Solve the following differential equation, given that

y = 1

when

x = 1

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

Solution

(a)

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

is homogeneous. Put

y = v x

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

\int \frac{2 v}{1 + v ^{2}} d v = - \int \frac{d x}{x} \Rightarrow lo g (1 + v^{2}) + lo g x = lo g C

(1 + v^{2}) x = C \Rightarrow x^{2} + y^{2} = C x

. With

y = 1, x = 1

C = 2

x^{2} + y^{2} = 2 x

(b)

\frac{d y}{d x} = \frac{2 y - x}{2 y + x}

is homogeneous. Put

y = v x

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

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

Splitting

2 v + 1 = \frac{1}{2} (4 v - 1) + \frac{3}{2}

and completing the square

2 v^{2} - v + 1 = 2 [(v - \frac{1}{4})^{2} + \frac{7}{16}]

lo g (2 v^{2} - v + 1) + \frac{6}{7} tan^{- 1} \frac{4 v - 1}{7} = - 2 lo g x + k^{'}

Put

v = \frac{y}{x}

and apply

y = 1, x = 1

to get

k^{'} = lo g 2 + \frac{6}{7} tan^{- 1} \frac{3}{7}

lo g (\frac{2 y ^{2} - x y + x ^{2}}{x ^{2}}) + \frac{6}{7} tan^{- 1} \frac{4 y - x}{7 x} + 2 lo g x = lo g 2 + \frac{6}{7} tan^{- 1} \frac{3}{7}

CBSE Class 12 Board · 2008 · Subjective

CBSE Class 12 · 2008 · Differential EquationsLong Answer

Solution

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