Question

The slope of the tangent to the curve at any point is equal to $y+2x$. Find the equation of the curve passing through the origin.

• A. $y+2(x+1)=2e^x$
• B. $y-2(x+1)=2e^x$
• C. $y+2(x+1)=e^x$
• D. $y-2(x+1)=e^x$

Answer 1

The correct option is A $y+2(x+1)=2e^x$

Slope of the tangent $=\frac{dy}{dx}$

According to the question, we have

$\frac{dy}{dx}=y+2x$

$\Rightarrow \frac{dy}{dx}-y=2x$ ... (i)

This is a linear differential equation.

Here, $P=-1$

$Q=2x$

$\therefore I.F.=e^{\int Pdx}=e^{\int (-1)dx}=e^{-x}$

$\therefore$ Solution of (i) is

$y(I.F.)=\int Q(I.F.)dx+c$, where $c$ is an arbitrary constant of integration.

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

$=2\int x{e}^{-xdx+c}$

$=2\left\{x\right(-e^{-x})-\int 1\cdot (-e^{-x}\left)dx\right\}+c$ [integrating by parts]

$=2[-x{e}^{-x}-e^{-x}]+c$

$=-2e^{-x}(x+1)+c$

$\Rightarrow y=-2e^{-x}(x+1)+c{e}^x$

If this curve passes through the origin,

$0=-2(0+1)+c{e}^0$

$\Rightarrow 0=-2+c$

$\Rightarrow c=2$

Hence, the equation of the curve is

$y=-2(x+1)+2e^x$

$\Rightarrow y=2(e^x-x-1)$.
$\Rightarrow y+2(x+1)=2e^x$