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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Solve the differential equation $\Large \frac{dy}{dx}\normalsize=1+x+y^2+xy^2,$when y=0,x=0.

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Toolbox:
  • A linear differential equation of the form.
  • $ \large\frac{dy}{dx}$$=f(x),$ can be solved by seperating the variable and then integrating it.
  • $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$\tan ^{-1} \bigg(\large\frac{x}{a}\bigg)$$+c$
Given $\large\frac{dy}{dx}$$=1+x+y^2+xy^2$ when $y=0,x=0$
This can be written as $\large\frac{dy}{dx}$$=(1+x)+y^2(1+x)$
=>$ \large\frac{dy}{dx}$$=(1+x)(1+y^2)$
Now seperating the variables we get,
$\large\frac{dy}{1+y^2}$$=dx(1+x)$
=>$\large\frac{dy}{1+y^2}$$=(1+x)dx$
Now integrating on both sides we get,
$\int \large\frac{dy}{1+y^2}$ is in the form of $\int \large\frac{dx}{x^2+a^2}=\frac{1}{a}$$ \tan ^{-1} (\large\frac{x}{a})+c$
$\tan ^{-1} \bigg(\large\frac{y}{1}\bigg)$$=x+\large\frac{x^2}{2}$$+c$
$\tan ^{-1} y=x+\large\frac{x^2}{2}$$+c$
Let us find the value of c, by substituting the given value of x and y
$\tan ^{-1}(0)=0+0+c$
But $\tan ^{-1} (0)=0$
Therefore $c=0$
Hence the required solution is
$\tan ^{-1}y =x+\large\frac{x^2}{2}$$=0$
$y =\tan \bigg(x+\large\frac{x^2}{2}\bigg)$
answered May 16, 2013 by meena.p
 
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