Y -xdydx -b 1-x 2 dydx

We #160;have,y xdydx=b1+x2dydx #8658;y b=bx2+xdydx #8658;1y bdy=1bx2+xdxIntegrating #160;both #160;sides, #160;we #160;get #8747;1y bdy= #8747 ...

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Byju's Answer

Standard XII

Mathematics

Factorization

y -xdydx =b 1...

Question

$y - x \frac{d y}{d x} = b (1 + x^{2} \frac{d y}{d x})$

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Solution

$We have, y - x \frac{d y}{d x} = b (1 + x^{2} \frac{d y}{d x}) \Rightarrow y - b = (b x^{2} + x) \frac{d y}{d x} \Rightarrow (\frac{1}{y - b}) d y = (\frac{1}{b x^{2} + x}) d x Integrating both sides, we get \int (\frac{1}{y - b}) d y = \int (\frac{1}{b x^{2} + x}) d x \Rightarrow \int (\frac{1}{y - b}) d y = \frac{1}{b} \int (\frac{1}{x^{2} + \frac{1}{b} x}) d x \Rightarrow \int (\frac{1}{y - b}) d y = \frac{1}{b} \int (\frac{1}{x^{2} + \frac{1}{b} x + \frac{1}{4 b^{2}} - \frac{1}{4 b^{2}}}) d x \Rightarrow \int (\frac{1}{y - b}) d y = \frac{1}{b} \int \frac{1}{{(x + \frac{1}{2 b})}^{2} - {(\frac{1}{2 b})}^{2}} d x \Rightarrow \log |y - b| = \frac{1}{2 \times \frac{1}{2 b} b} \log |\frac{x + \frac{1}{2 b} - \frac{1}{2 b}}{x + \frac{1}{2 b} + \frac{1}{2 b}}| + \log C \Rightarrow \log |y - b| = \log |\frac{b x}{b x + 1}| + \log C \Rightarrow y - b = \frac{C b x}{b x + 1} \Rightarrow C b x = (y - b) (b x + 1) \Rightarrow x = k (y - b) (b x + 1), where k = \frac{1}{b C}$

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Similar questions

Q. Find one-parameter families of solution curves of the following differential equations:
(or Solve the following differential equations)
(i)

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

, m is a given real number

(ii)

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

(iii)

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

(iv)

x \frac{d y}{d x} + y = x^{4}

(v)

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

(vi)

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

(vii)

\frac{d y}{d x} + y \cos x = e^{\sin x} \cos x

(viii)

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

(ix)

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

(x)

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

(xi)

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

(xii)

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

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}

Q. Solve the each of the following differential equations:
(i)

(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

(iv)

\frac{d y}{d x} - y = \cos x

(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}

(viii)

\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. The solution of

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

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}

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y-xdydx=b1+x2dydx

$y - x \frac{d y}{d x} = b (1 + x^{2} \frac{d y}{d x})$