dydx=y+xx ⇒dydx yx=1 Integrating factor =e^∫ 1xdx=e^ log x I.F=1x Then solution, =I.F.y=∫ I.F dx ⇒ yx=∫1xdx ⇒ yx=log x+c ...

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Standard XII

Mathematics

Linear Differential Equations of First Order

Solve: dydx...

Question

Solve: $\frac{d y}{d x} = \frac{y + x}{x}$

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Solution

$\frac{d y}{d x} = \frac{y + x}{x}$

$\Rightarrow \frac{d y}{d x} - \frac{y}{x} = 1$

Integrating factor $= e^{\int - \frac{1}{x} d x} = e^{- l o g x}$

$I . F = \frac{1}{x}$

Then solution, $= I . F . y = \int I . F d x$

$\Rightarrow y x = \int \frac{1}{x} d x$

$\Rightarrow y x = l o g x + c$

$\Rightarrow y = \frac{l o g x + c}{x}$ .

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\frac{d y}{d x} + y \cot x = 4 x cosec x, y (\frac{π}{2}) = 0

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Solve: dydx=y+xx

dydx=y+xx⇒dydx−yx=1Integrating factor =e∫−1xdx=e−logxI.F=1xThen solution, =I.F.y=∫I.Fdx⇒yx=∫1xdx⇒yx=logx+c⇒y=logx+cx.

Solve: $\frac{d y}{d x} = \frac{y + x}{x}$