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Differential Equations question

2023 · 13 Apr · Shift 1 · Q27

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Mathematics
Differential Equations
2023 · 13 Apr · Shift 1 · Q27

Differential Equations question

2023 · 13 Apr · Shift 1 · Q27

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

Let

y=y_{1}(x)

and

y=y_{2}(x)

be the solution curves of the differential equation

\frac{d y}{d x}=y+7

with initial conditions

y_{1}(0)=0

and

y_{2}(0)=1

respectively. Then the curves

y=y_{1}(x)

and

y=y_{2}(x)

intersect at

A
no point
B
two points
C
infinite number of points
D
one point

View written solutionFree

Correct answer: A

Solve the differential equation

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

Rewrite it as $\frac{dy}{dx}-y=7.$

This is a first-order linear differential equation.

Find the general solution

We can also solve by shifting variable. Let $u=y+7.$ Then $\frac{du}{dx}=\frac{dy}{dx}=y+7=u.$

So, $\frac{du}{dx}=u \implies u=Ce^x,$ where $C$ is a constant.

Hence, $y+7=Ce^x$ so the general solution is $y=Ce^x-7.$

Use the initial conditions

For $y_1(x)$ :

Given $y_1(0)=0$ , $0=C\cdot e^0-7=C-7,$ so $C=7.$ Thus, $y_1(x)=7e^x-7.$

For $y_2(x)$ :

Given $y_2(0)=1$ , $1=C\cdot e^0-7=C-7,$ so $C=8.$ Thus, $y_2(x)=8e^x-7.$

Check whether the curves intersect

For intersection, we need $y_1(x)=y_2(x).$

So, $7e^x-7=8e^x-7.$ Cancel $-7$ from both sides: $7e^x=8e^x.$ Thus, $e^x=0.$

But this is impossible since $e^x>0 \quad \text{for all real } x.$

Therefore, the two curves never intersect.

Evaluate options

A: no point — Correct
B: two points — Incorrect
C: infinite number of points — Incorrect
D: one point — Incorrect

So the correct answer is A.

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