If xdydx-x2-y-2- y1-1- and x-0 then y2 equals

Given dydx 1xy=(x 2x) I.F.=e^ ∫1xdx=e^ ln|x|=1x Now general solution is given by yx=∫(x 2x)1xdx ⇒yx=x+2x+C As y(1)=1⇒C= 2 ∴yx=x+2x 2 ⇒y=x^2 2x+2 Hence,y(2)=(2)^ ...

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Standard XII

Mathematics

General Solution of a Differential Equation

If xdydx=x2+y...

Question

If $x \frac{d y}{d x} = x^{2} + y - 2, y (1) = 1,$ and $x > 0$ then $y (2)$ equals

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Solution

Given $\frac{d y}{d x} - \frac{1}{x} y = (x - \frac{2}{x})$
I.F. $= e^{- \int \frac{1}{x} d x} = e^{- ln | x |} = \frac{1}{x}$
Now general solution is given by $\frac{y}{x} = \int (x - \frac{2}{x}) \frac{1}{x} d x$
$\Rightarrow \frac{y}{x} = x + \frac{2}{x} + C$
As $y (1) = 1 \Rightarrow C = - 2$
$∴ \frac{y}{x} = x + \frac{2}{x} - 2 \Rightarrow y = x^{2} - 2 x + 2$
Hence, $y (2) = {(2)}^{2} - 2 (2) + 2 = 2$

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General Solution of a Differential Equation

Standard XII Mathematics

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If xdydx=x2+y−2, y(1)=1, and x>0 then y(2) equals

Given dydx−1xy=(x−2x) I.F.=e−∫1xdx=e−ln|x|=1x Now general solution is given by yx=∫(x−2x)1xdx ⇒yx=x+2x+C As y(1)=1⇒C=−2 ∴yx=x+2x−2⇒y=x2−2x+2 Hence, y(2)=(2)2−2(2)+2=2

If $x \frac{d y}{d x} = x^{2} + y - 2, y (1) = 1,$ and $x > 0$ then $y (2)$ equals