JEE Mathematics 2023 Question with Solution

Given: $\frac{dy}{dx} = y + 7$ with two initial conditions $y_1(0)=0$ and $y_2(0)=1$.

Find: Whether the curves $y=y_1(x)$ and $y=y_2(x)$ intersect.

Solve the differential equation by separation of variables:

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

Integrating,

$$\int \frac{1}{y+7}\,dy = \int 1\,dx$$

So,

$$\ln|y+7| = x + C$$

Hence,

$$y = -7 + Ae^x$$

Now apply the initial conditions.

For $y_1(0)=0$,

$$0 = -7 + A$$

Therefore,

$$A=7$$

and

$$y_1(x) = -7 + 7e^x$$

For $y_2(0)=1$,

$$1 = -7 + A$$

Therefore,

$$A=8$$

and

$$y_2(x) = -7 + 8e^x$$

For intersection, set

$$y_1(x)=y_2(x)$$

So,

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

which gives

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

and hence

$$e^x=0$$

But $e^x$ is never zero for any real $x$. Therefore, the two curves do not intersect at any point.

The correct option is A.

Given: Both curves satisfy the same differential equation $\frac{dy}{dx}=y+7$.

Find: Whether they can meet.

The general solution is

$$y = -7 + Ae^x$$

The two initial conditions give different constants: $A=7$ and $A=8$. Thus the two curves are

$$y_1(x)=-7+7e^x, \qquad y_2(x)=-7+8e^x$$

Their difference is

$$y_2(x)-y_1(x)=e^x$$

Since $e^x>0$ for every real $x$, the difference is never zero. Therefore the curves never intersect.

The correct option is A.