The solution of dydx-x-y-12x-2y-3 is-

Here nbsp;a1a2=b1b2Let nbsp;x+y=v1+dydx=dvdxdydx=dvdx 1dvdx 1=v+12v+3dvdx=3v+42v+3∫2v+33v+4dv=∫dx nbsp;+c23∫dv nbsp;+ nbsp;13∫13v+4dv nbsp;=x+c23v ...

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Standard XIII

Mathematics

Equations Reducible to Standard Forms

The solution ...

Question

$The solution of \frac{dy}{dx} = \frac{x + y + 1}{2 x + 2 y + 3} is$ .

\frac{2}{3} (x + y) + \frac{1}{9} log (2 x + 2 y + 4 | = x + c

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\frac{2}{3} (x + y) + \frac{1}{9} log (3 x + 3 y + 4 | = x + c

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\frac{2}{3} (x + y) + \frac{1}{9} log (3 x + 3 y | = x + c

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\frac{1}{9} log (3 x + 3 y + 4 | = x + c

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Solution

The correct option is B $\frac{2}{3} (x + y) + \frac{1}{9} log (3 x + 3 y + 4 | = x + c$
$Here \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} Let x + y = v 1 + \frac{dy}{dx} = \frac{dv}{dx} \frac{dy}{dx} = \frac{dv}{dx} - 1 \frac{dv}{dx} - 1 = \frac{v + 1}{2 v + 3} \frac{dv}{dx} = \frac{3 v + 4}{2 v + 3} \int \frac{2 v + 3}{3 v + 4} dv = \int dx + c \frac{2}{3} \int dv + \frac{1}{3} \int \frac{1}{3 v + 4} dv = x + c \frac{2}{3} v + \frac{1}{9} log (3 v + 4 | = x + c \frac{2}{3} (x + y) + \frac{1}{9} log (3 x + 3 y + 4 | = x + c$

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Q. Show that y = ae^2x + be^−x is a solution of the differential equation

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

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

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The solution of dydx=x+y+12x+2y+3 is.

The correct option is B 23(x+y)+19log(3x+3y+4|=x+cHere a1a2=b1b2Let x+y=v1+dydx=dvdxdydx=dvdx−1dvdx−1=v+12v+3dvdx=3v+42v+3∫2v+33v+4dv=∫dx +c23∫dv + 13∫13v+4dv =x+c23v +19log(3v+4|=x+c23(x+y)+19log(3x+3y+4|=x+c

$The solution of \frac{dy}{dx} = \frac{x + y + 1}{2 x + 2 y + 3} is$ .