Question

Find the equation of the curve passing through the point$\left(0,1\right)$, if the slope of tangent to the curve at each of its point is equal to the sum of the abscissa and the product of the abscissa and ordinate of the point.

Answer 1

According to the question,

$\frac{dy}{dx}=x+xy$

$\Rightarrow \frac{dy}{dx}-xy=x$

Comparing with $\frac{dy}{dx}+Py=Q$, we get

$P=-x,Q=x$

Now,$I.F=e^{-\int xdx}=e^{-\frac{x^2}{2}}$

So, the solution is given by

$y\times I.F=\int Q\times I.Fdx+c$

$\Rightarrow y{e}^{-\frac{x^2}{2}}=\int x{e}^{-\frac{x^2}{2}}dx+c$

Putting $-\frac{x^2}{2}=t\Rightarrow -xdx=dt$

$\therefore I=-\int e^{t}dt=-e^t+c$

$\therefore I=-e^{-\frac{x^2}{2}}+c$

$\therefore y{e}^{-\frac{x^2}{2}}=-e^{-\frac{x^2}{2}}+c$

Since the curve passes through the point $(0,1),$ it satisfies the equation of the curve.

$\Rightarrow 1e^0=-e^0+c$

$\Rightarrow c=1+1=2$

Putting the value of $C$ in the equation of the curve, we get

$y{e}^{-\frac{x^2}{2}}=-e^{-\frac{x^2}{2}}+2$

or $y=-1+2e^{-\frac{x^2}{2}}$