JEE Main 2023 — Differential Equations Question with Solution

The given differential equation is

$$\frac{dy}{dx}=y+7$$

This is a first order linear differential equation and can be solved using an integrating factor.

Rearrange the equation to the standard form of a linear differential equation :

$$\frac{dy}{dx}-y=7$$

The integrating factor is $e^{-\int dx}=e^{-x}$.

Multiplying each side of the equation by the integrating factor gives :

$$e^{-x}\frac{dy}{dx}-e^{-x}y=7e^{-x}$$

The left-hand side of the equation is the derivative of $(e^{-x}y)$ with respect to $x$. So we can write the equation as :

$$\frac{d}{dx}(e^{-x}y)=7e^{-x}$$

Integrate both sides with respect to $x$ :

$$e^{-x}y=-7e^{-x}+C$$

Multiply both sides by $e^x$ to isolate $y$ :

$$y=-7+C{e}^x$$

So, the general solution to the differential equation is $y=-7+C{e}^x$.

Now, let's apply the initial conditions to find the particular solutions :

For $y_1(0)=0$, we substitute into the general solution and solve for $C$ :

$$0=-7+C$$

So, $C=7$, and the solution for $y_1$ is $y_1(x)=7e^x-7$.

For $y_2(0)=1$, again substitute into the general solution:

$$1=-7+C$$

So, $C=8$, and the solution for $y_2$ is $y_2(x)=8e^x-7$.

The two curves intersect when $y_1(x)=y_2(x)$. Setting these equal and solving for $x$ gives :

$$7e^x-7=8e^x-7$$

$$e^x=0$$

But $e^x=0$ has no solution, because the exponential function never equals zero.

So, the curves $y=y_1(x)$ and $y=y_2(x)$ do not intersect at any point.