The solution of dydx- x - y -12 x -2 y -3 is-A- 23 x-y-19 log 2 x-2 y-4-x-cB- 23 x-y-19 log 3 x-3 y-4-x-cC- 23 x-y-19 log 3 x-3 y-x-cD- 19 log 3 x-3 y-4-x-c

The correct option is B 23x+y+19log3x+3y+4=x+cHere nbsp;a1a2=b1b2Let nbsp;x+y=v1+dydx=dvdxdydx=dvdx 1dvdx 1=v+12v+3dvdx=3v+42v+3∫2v+33v+4dv=∫dx nbsp;+c23∫ ...

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Byju's Answer

Standard XII

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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Similar questions

Q. Solve the system of equations

x + 2 y + 3 z = 1 2 x + 3 y + 2 z = 2 3 x + 3 y + 4 z = 1

Find the value of

14 (x + y + z)

Q. If the system of linear equations

\begin{matrix} x - 2 y + k z = 1 2 x + y + z = 2 3 x - y - k z = 3 \end{matrix}

has a solution

(x, y, z), z \neq 0

, then

(x, y)

lies on the straight line whose equation is :

1/3y+ y + 1/3x-y=3/4

1/2(3x+y)-1/2(3x-y)=-1/8

Solve for x and y

Q. Solve the following pairs of equations by reducing them to a pair of linear equations:

(i)

\frac{1}{2 x} + \frac{1}{3 y} = 2; \frac{1}{3 x} + \frac{1}{2 y} = \frac{13}{6}

(ii)

\frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2; \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = - 1

(iii)

\frac{4}{x} + 3 y = 14; \frac{3}{x} - 4 y = 23

(iv)

\frac{5}{x - 1} + \frac{1}{y - 2} = 2; \frac{6}{x - 1} - \frac{3}{y - 2} = 1

(v)

\frac{7 x - 2 y}{x y} = 5; \frac{8 x + 7 y}{x y} = 15

(vi)

6 x + 3 y = 6 x y; 2 x + 4 y = 5 x y

(vii)

\frac{10}{x + y} + \frac{2}{x - y} = 4; \frac{15}{x + y} - \frac{5}{x - y} = - 2

(viii)

\frac{1}{3 x + y} + \frac{1}{3 x - y} = \frac{3}{4}; \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = \frac{- 1}{8}

Q. Evaluate :

\frac{1}{3 x + y} + \frac{1}{3 y - y} = \frac{3}{4}; \frac{1}{2 (3 x + y)} - \frac{1}{2 (3 x - y)} = - \frac{1}{8}

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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

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