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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Differential Equations

General solution of differential equation: $\large\frac{dy}{dx} =\frac{2x+3y+6}{4x+6y+7}$

$(a)\;7x+6y-15 \log(7(4x+6y+7)+15)=c \\ (b)\;2x+3y-15 \log(7(2x+3y-15)+15)=c \\ (c)\;-3x+6y-15 \log(7(4x+6y+7)+15)=c \\ (d)\;-x+3y-15 \log(7(2x+3y-15)+15)=c $

1 Answer

$4x+6y+7=t$
$2(2x+3y)+7=t$
$4 +\large\frac{6 dy}{dx} =\frac{dt}{dx}$
$\large\frac{1}{6} \frac{dt}{dx}-\frac{4}{6}=\large\frac{\Large\frac{(t-7)}{2}+6}{t}$
$\large\frac{dt}{dx}=\frac{3 (t-7)+36+4t}{t}$
$\large\frac{1}{7} \large\frac{t dt}{t+ 15/7}$$=dx$
$\large\frac{1}{7} \int \large\frac{t+\Large\frac{15}{7}}{t+ \Large\frac{15}{7}}- \large\frac{\Large\frac{15}{7}}{t+ \Large\frac{15}{7}} $$dt=\int dt$
$\large\frac{1}{7} t -\frac{15}{7} $$\log \bigg(t+\large\frac{15}{7} \bigg)=7x+c''$
$ -3x+6y-15 \log(7(4x+6y+7)+15)=c$
Hence c is the correct answer

 

answered Feb 4, 2014 by meena.p
 

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