We #160;have,dydx+1+y2y=0 #8658;dydx= 1+y2y #8658;dxdy= y1+y2 #8658;dx= y1+y2dyIntegrating #160;both #160;sides, #160;we #160;get #8747;dx= ...

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Standard VII

Mathematics

Algebra Generalized Arithmetic

dydx +1+ y 2 ...

Question

$\frac{d y}{d x} + \frac{1 + y^{2}}{y} = 0$

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Solution

$We have, \frac{d y}{d x} + \frac{1 + y^{2}}{y} = 0$
$\Rightarrow \frac{d y}{d x} = - \frac{(1 + y^{2})}{y} \Rightarrow \frac{d x}{d y} = - \frac{y}{1 + y^{2}} \Rightarrow d x = (- \frac{y}{1 + y^{2}}) d y Integrating both sides, we get \int d x = \int (- \frac{y}{1 + y^{2}}) d y \Rightarrow x = \int (- \frac{y}{1 + y^{2}}) d y Putting 1 + y^{2} = t, we get 2 y d y = d t ∴ x = - \frac{1}{2} \int \frac{1}{t} d t \Rightarrow x = - \frac{1}{2} \log |t| + C \Rightarrow x = - \frac{1}{2} \log |1 + y^{2}| + C \Rightarrow x + \frac{1}{2} \log |1 + y^{2}| = C Hence, x + \frac{1}{2} \log |1 + y^{2}| = C is the required solution .$

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Similar questions

Q. Differential equation

\frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + 2 y = 0, y (0) = 1, y' (0) = 3

Function y = e^x + e^2x

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]

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

(ii)

x \cos (\frac{y}{x}) \frac{d y}{d x} = y \cos (\frac{y}{x}) + x

(iii) y dx + x log

(\frac{y}{x})

dy − 2x dy = 0

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(v)

x \frac{d y}{d x} + 2 y = x^{2}, x \neq 0

(vi)

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

(vii)

\frac{d y}{d x} + 3 y = e^{- 2 x}

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

(ix)

\frac{d y}{d x} + (\sec x) y = \tan x

(x)

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

(xi)

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

(xii) (1 + x²) dy + 2xy dx = cot x dx

(xiii)

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

(xiv) y dx + (x − y²) dy = 0

(xv)

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

Q. If y = ae^2x + be^−x, show that,

\frac{d^{2} y}{d x^{2}} - \frac{d y}{d x} - 2 y = 0

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

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(x)

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

[CBSE 2018]

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dydx+1+y2y=0

We have,dydx+1+y2y=0 ⇒dydx=-1+y2y⇒dxdy=-y1+y2⇒dx=-y1+y2dyIntegrating both sides, we get∫dx=∫-y1+y2dy⇒x=∫-y1+y2dyPutting 1+y2=t, we get2y dy=dt∴x=-12∫1tdt⇒x=-12logt+C⇒x=-12log1+y2+C⇒x+12log1+y2=CHence, x+12log1+y2=C is the required solution.

$\frac{d y}{d x} + \frac{1 + y^{2}}{y} = 0$