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

Find general solution of equation $y'-2y+a=0$ a=fixed

$(a)\;y=ce^{2x}+\frac{a}{2} \\ (b)\;y= e^{3x}+c \\ (c)\;y= e^{2x}+K \\ (d)\;y=e^{-x}+K $

1 Answer

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$\large\frac{dy}{2y-a}$$=dx$ for $ y \neq \large\frac{a}{2}$
For $y= \large\frac{a}{2}$ we find $\large\frac{dy}{dx}$$=0$
and differential equation is satisfied .
Hence $y= \large\frac{a}{2}$ is solution. Since it does not contain an arbitrary constant so it s not general solution .
$\large\frac{1}{2} $$ \log (2y-a) =x+c_1$
or $2y-a=e^{2x+2C_1}$
$2y-a =Ke^{2x}$
$K= e \pm e^{2x}$ is arbitrary constant
$c =\large\frac{K}{2}$
$y= Ce^{2x}+\large\frac{a}{2}$
Hence a is the correct answer.
answered Feb 7, 2014 by meena.p

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