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Byju's Answer

Standard XII

Mathematics

Standard Formulae - 3

i d y d x=y t...

Question

(i) $\frac{d y}{d x} = y \tan x, y (0) = 1$

(ii) $2 x \frac{d y}{d x} = 5 y, y (1) = 1$

(iii) $\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1$

(iv) $\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}$

(v) $\frac{d y}{d x} = 2 x y, y (0) = 1$

(vi) $\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1$

(vii) $x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1$
(viii) $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$ when y = 0, x = 0 [NCERT EXEMPLAR]
(ix) $2 (y + 3) - x y \frac{d y}{d x} = 0$ , y(1) = −2 [NCERT EXEMPLAR]

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Solution

$(i) \frac{d y}{d x} = y \tan x, y (0) = 1 \Rightarrow \frac{1}{y} d y = \tan x d x Integrating both sides, we get \int \frac{1}{y} d y = \int \tan x d x \Rightarrow \log |y| = \log |\sec x| + C . . . . . (1) We know that at x = 0 and y = 1 . Substituting the values of x and y in (1), we get \log |1| = \log |1| + C \Rightarrow C = 0 Substituting the value of C in (1), we get$

$\log |y| = \log |\sec x| + 0 \Rightarrow y = \sec x Hence, y = \sec x, where x \in (\frac{- π}{2}, \frac{π}{2}), is the required solution .$

$(ii) 2 x \frac{d y}{d x} = 5 y, y (1) = 1 \Rightarrow \frac{2}{y} d y = \frac{5}{x} d x Integrating both sides, we get 2 \int \frac{1}{y} d y = 5 \int \frac{1}{x} d x \Rightarrow 2 \log |y| = 5 \log |x| + C . . . . . (1) We know that at x = 1 and y = 1 . Substituting the values of x and y in (1), we get 2 \log |1| = 5 \log |1| + C \Rightarrow C = 0 Substituting the value of C in (1), we get$

$2 \log |y| = 5 \log |x| + 0 \Rightarrow y = {|x|}^{\frac{5}{2}} Hence, y = {|x|}^{\frac{5}{2}} is the required solution .$

$(iii) \frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1 \Rightarrow \frac{1}{y^{2}} d y = 2 e^{2 x} d x Integrating both sides, we get \int \frac{1}{y^{2}} d y = 2 \int e^{2 x} d x \Rightarrow \frac{- 1}{y} = e^{2 x} + C . . . . . (1) We know that at x = 0, y = - 1 . Substituting the values of x and y in (1), we get 1 = 1 + C \Rightarrow C = 0 Substituting the value of C in (1), we get$

$- \frac{1}{y} = e^{2 x} \Rightarrow y = - e^{- 2 x} Hence, y = - e^{- 2 x} is the required solution .$

$(iv) \cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2} \Rightarrow \cos y d y = e^{x} d x Integrating both sides, we get \int \cos y d y = \int e^{x} d x \Rightarrow \sin y = e^{x} + C . . . . . (1) We know that at x = 0, y = \frac{π}{2} . Substituting the values of x and y in (1), we get 1 = 1 + C \Rightarrow C = 0 Substituting the value of C in (1), we get$

$\sin y = e^{x} \Rightarrow y = \sin^{- 1} (e^{x}) Hence, y = \sin^{- 1} (e^{x}) is the required solution .$

$(v) \frac{d y}{d x} = 2 x y, y (0) = 1 \Rightarrow \frac{1}{y} d y = 2 x d x Integrating both sides, we get \int \frac{1}{y} d y = \int 2 x d x \log |y| = x^{2} + C . . . . . (1) We know that at x = 0, y = 1 . Substituting the values of x and y in (1), we get 0 = 0 + C \Rightarrow C = 0 Substituting the value of C in (1), we get \log |y| = x^{2} \Rightarrow y = e^{x^{2}} Hence, y = e^{x^{2}} is the required solution .$

$(vi) \frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1 \Rightarrow \frac{d y}{d x} = (1 + x^{2}) (1 + y^{2}) \Rightarrow \frac{d y}{(1 + y^{2})} = (1 + x^{2}) d x Integrating both sides, we get \int \frac{d y}{(1 + y^{2})} = \int (1 + x^{2}) d x \Rightarrow \tan -^{1} y = x + \frac{x^{3}}{3} + C . . . . . (1) We know that at x = 0, y = 1 . Substituting the values of x and y in (1), we get \frac{π}{4} = 0 + 0 + C \Rightarrow C = \frac{π}{4} Substituting the value of C in (1), we get \tan -^{1} y = x + \frac{x^{3}}{3} + \frac{π}{4} Hence, \tan -^{1} y = x + \frac{x^{3}}{3} + \frac{π}{4} is the required solution .$

$(vii) x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1 \Rightarrow \frac{y}{y + 2} d y = \frac{x + 2}{x} d x \Rightarrow \frac{y + 2 - 2}{y + 2} d y = \frac{x + 2}{x} d x \Rightarrow (1 - \frac{2}{y + 2}) d y = (1 + \frac{2}{x}) d x Integrating both sides, we get \int (1 - \frac{2}{y + 2}) d y = \int (1 + \frac{2}{x}) d x \Rightarrow y - 2 \log |y + 2| = x + 2 \log |x| + C . . . . . (1) We know that at x = 1, y = - 1 . Substituting the values of x and y in (1), we get - 1 - 2 \log |1| = 1 + 2 \log |1| + C \Rightarrow - 1 = 1 + C \Rightarrow C = - 2 Substituting the value of C in (1), we get y - 2 \log |y + 2| = x + 2 \log |x| - 2 Hence, y - 2 \log |y + 2| = x + 2 \log |x| - 2 is the required solution .$
(viii) $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$
$\Rightarrow \frac{d y}{d x} = 1 + x + y^{2} (1 + x) \Rightarrow \frac{d y}{d x} = (1 + x) (1 + y^{2}) \Rightarrow \frac{d y}{(1 + y^{2})} = (1 + x) d x \Rightarrow \int \frac{d y}{(1 + y^{2})} = \int (1 + x) d x \Rightarrow \tan^{- 1} y = x + \frac{x^{2}}{2} + C . . . . . (1) Now, \tan^{- 1} 0 = 0 + 0 + C [∵ y = 0, x = 0] \Rightarrow C = 0 Substituting the value of C in (1), we get \tan^{- 1} y = x + \frac{x^{2}}{2} \Rightarrow y = \tan (x + \frac{x^{2}}{2})$

(ix) $2 (y + 3) - x y \frac{d y}{d x} = 0$
$\Rightarrow 2 (y + 3) = x y \frac{d y}{d x} \Rightarrow \frac{2}{x} d x = \frac{y}{y + 3} d y \Rightarrow \frac{2}{x} d x = \frac{y + 3 - 3}{y + 3} d y \Rightarrow \frac{2}{x} d x = (1 - \frac{3}{y + 3}) d y \Rightarrow \int \frac{2}{x} d x = \int (1 - \frac{3}{y + 3}) d y \Rightarrow 2 \log x = y - 3 \log |y + 3| + C \Rightarrow \log x^{2} + \log |{(y + 3)}^{3}| = y + C \Rightarrow \log |(x^{2}) {(y + 3)}^{3}| = y + C . . . . . (1)$
$\Rightarrow \log |{(1)}^{2} {(- 2 + 3)}^{3}| = - 2 + C \Rightarrow C = 2 Substituting the value of C in (1), we get \log |(x^{2}) {(y + 3)}^{3}| = y + 2 \Rightarrow (x^{2}) {(y + 3)}^{3} = e^{y + 2}$

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Similar questions

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

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

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

(viii)

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

when y = 0, x = 0 [NCERT EXEMPLAR]
(ix)

2 (y + 3) - x y \frac{d y}{d x} = 0

, y(1) = −2 [NCERT EXEMPLAR]
(x)

e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0, y (0) = \frac{π}{4}

[CBSE 2018]

Q. (i)

\frac{d y}{d x} = y \tan x, y (0) = 1

(ii)

2 x \frac{d y}{d x} = 5 y, y (1) = 1

(iii)

\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1

(iv)

\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}

(v)

\frac{d y}{d x} = 2 x y, y (0) = 1

(vi)

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

(vii)

x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1

Q. Solve each of the following initial value problems:
(i)

y' + y = e^{x}, y (0) = \frac{1}{2}

(ii)

x \frac{d y}{d x} - y = \log x, y (1) = 0

(iii)

\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0

(iv)

x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0

(v)

(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0

(vi)

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1

(vii)

x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1

(viii)

\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

(ix)

\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}

(x)

\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}

Q. Solve each of the following initial value problems:
(i)

y' + y = e^{x}, y (0) = \frac{1}{2}

(ii)

x \frac{d y}{d x} - y = \log x, y (1) = 0

(iii)

\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0

(iv)

x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0

(v)

(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0

(vi)

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1

(vii)

x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1

(viii)

\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

(ix)

\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}

(x)

\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}

(xi)
(xii)

Q. Solve each of the following initial value problems:
(i)

y' + y = e^{x}, y (0) = \frac{1}{2}

(ii)

x \frac{d y}{d x} - y = \log x, y (1) = 0

(iii)

\frac{d y}{d x} + 2 y = e^{- 2 x} \sin x, y (0) = 0

(iv)

x \frac{d y}{d x} - y = (x + 1) e^{- x}, y (1) = 0

(v)

(1 + y^{2}) d x + (x - e^{- \tan^{- 1} y}) d x = 0, y (0) = 0

(vi)

\frac{d y}{d x} + y \tan x = 2 x + x^{2} \tan x, y (0) = 1

(vii)

x \frac{d y}{d x} + y = x \cos x + \sin x, y (\frac{π}{2}) = 1

(viii)

\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

(ix)

\frac{d y}{d x} + 2 y \tan x = \sin x; y = 0 when x = \frac{π}{3}

(x)

\frac{d y}{d x} - 3 y \cot x = \sin 2 x; y = 2 when x = \frac{π}{2}

(xi)

\frac{d y}{d x} + y \cot x = 2 \cos x, y (\frac{π}{2}) = 0

(xii)

d y = \cos x (2 - y \cos e c x) d x

(xiii)

\tan x (\frac{d y}{d x}) = 2 x \tan x + x^{2} - y; \tan x \neq 0

given that y = 0 when

x = \frac{π}{2}

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(i) dydx=y tan x, y0=1 (ii) 2xdydx=5y, y1=1 (iii) dydx=2e2x y2, y0=-1 (iv) cos ydydx=ex, y0=π2 (v) dydx=2xy, y0=1 (vi) dydx=1+x2+y2+x2y2, y0=1 (vii) xydydx=x+2y+2, y1=-1 (viii) dydx=1+x+y2+xy2 when y = 0, x = 0 [NCERT EXEMPLAR] (ix) 2y+3-xydydx=0, y(1) = −2 [NCERT EXEMPLAR]