QJEEM23Ma762f: JEE Main 2023 Mathematics - Class 12 JEEM23Ma762f Solution: Let y = y 1 ( x ) and y = y 2 ( x ) be the solution curves of…

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 $y = -7 + Ce^{x}$ .

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.