Solve the differential equation d y-d x-1-x-y2-x y2- when y-0 and x-0-

Given that, dy/dx=1+x+y^2+xy^2 ⇒dy/dx=(1+x)+y^2(1+x) ⇒dy/dx=(1+y^2)(1+x) ⇒dy/1+y^2=(1+x)dx On integrating bothe sides, we get tan^ 1y=x+x^2/2+K ...(i) ...

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Standard XII

Mathematics

Solving Homogeneous Differential Equations

Solve the dif...

Question

Solve the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2},$ when y=0 and x=0.

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Solution

Given that, $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$
$\Rightarrow \frac{d y}{d x} = (1 + x) + y^{2} (1 + x) \Rightarrow \frac{d y}{d x} = (1 + y^{2}) (1 + x) \Rightarrow \frac{d y}{1 + y^{2}} = (1 + x) d x$
On integrating bothe sides, we get
$t a n^{- 1} y = x + \frac{x^{2}}{2} + K$ ...(i)
when y=0 and x=0, then substituting these values in Eq. (i), we get
$t a n^{- 1} (0) = 0 + 0 + K \Rightarrow K = 0 \Rightarrow t a n^{- 1} y = x + \frac{x^{2}}{2} \Rightarrow y = t a n (x + \frac{x^{2}}{2})$

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Solve the differential equation dydx=1+x+y2+xy2, when y=0 and x=0.

Given that, dydx=1+x+y2+xy2 ⇒dydx=(1+x)+y2(1+x)⇒dydx=(1+y2)(1+x)⇒dy1+y2=(1+x)dx On integrating bothe sides, we get tan−1y=x+x22+K ...(i) when y=0 and x=0, then substituting these values in Eq. (i), we get tan−1(0)=0+0+K⇒K=0⇒tan−1y=x+x22⇒y=tan(x+x22)

Solve the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2},$ when y=0 and x=0.