Question

If the slope of the tangent of the curve at any point is equal to $-y+e^{-x}$, then the equation of the curve passing through origin is

• A. $y+xe^x=0$
• B. $ye^x+x=0$
• C. $ye^x-x=0$
• D. $y-xe^x=0$

Hint:

This is a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The integrating factor (I.F.) is $e^{\int P(x)dx}$, and the solution is $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$.

Answer 1

The Correct Option is A

Step 1: The slope of the tangent to the curve at any point is given by $\frac{dy}{dx}$. So, the differential equation is:\n\[\frac{dy}{dx} = -y+e^{-x}\]
Step 2: Rearrange the equation into the standard form of a linear first-order differential equation, $\frac{dy}{dx} + P(x)y = Q(x)$.\n\[\frac{dy}{dx} + y = e^{-x}\] Here, $P(x) = 1$ and $Q(x) = e^{-x}$. Now, calculate the integrating factor (I.F.):\n\[\text{I.F.} = e^{\int P(x) dx} = e^{\int 1 dx} = e^x\]
Step 3: The general solution of a linear first-order differential equation is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$.\n\[y e^x = \int e^{-x} \cdot e^x dx + C\] \[y e^x = \int 1 dx + C\] \[y e^x = x + C \quad \ldots (i)\]
Step 4: The curve passes through the origin $(0,0)$. Substitute $x=0$ and $y=0$ into the general solution to find the constant $C$.\n\[(0) e^0 = 0 + C\] \[0 = 0 + C \implies C = 0\]
Step 5: Substitute the value of $C$ back into equation (i) to get the particular equation of the curve.\n\[y e^x = x + 0\] \[y e^x - x = 0\]