Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Differential Equations question

2023 · 25 Jan · Shift 2 · Q35
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Differential Equations
  5. /2023 · 25 Jan · Shift 2 · Q35

Differential Equations question

2023 · 25 Jan · Shift 2 · Q35

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(t)y=y(t)y=y(t) be a solution of the differential equation dydt+αy=γe−βt{{dy} \over {dt}} + \alpha y = \gamma {e^{ - \beta t}}dtdy​+αy=γe−βt where, α>0,β>0\alpha \gt 0,\beta \gt 0α>0,β>0 and γ>0\gamma \gt 0γ>0. Then lim⁡t→∞y(t)\mathop {\lim }\limits_{t \to \infty } y(t)t→∞lim​y(t)
  1. A
    is 0
  2. B
    is 1
  3. C
    is −1-1−1
  4. D
    does not exist
View written solutionFree

Correct answer: A

  1. We are given the linear differential equation
dydt+αy=γe−βt,\frac{dy}{dt}+\alpha y=\gamma e^{-\beta t},dtdy​+αy=γe−βt,

where α>0,β>0,γ>0\alpha>0,\beta>0,\gamma>0α>0,β>0,γ>0.

We need to find

lim⁡t→∞y(t).\lim_{t\to\infty} y(t).t→∞lim​y(t).
  1. Solve the differential equation using the integrating factor method.

The integrating factor is

I.F.=e∫α dt=eαt.I.F.=e^{\int \alpha \,dt}=e^{\alpha t}.I.F.=e∫αdt=eαt.

Multiplying the equation by eαte^{\alpha t}eαt,

eαtdydt+αeαty=γe(α−β)t.e^{\alpha t}\frac{dy}{dt}+\alpha e^{\alpha t}y=\gamma e^{(\alpha-\beta)t}.eαtdtdy​+αeαty=γe(α−β)t.

So,

ddt(yeαt)=γe(α−β)t.\frac{d}{dt}\left(ye^{\alpha t}\right)=\gamma e^{(\alpha-\beta)t}.dtd​(yeαt)=γe(α−β)t.
  1. Integrate.

There are two cases.

Case 1: α≠β\alpha\neq \betaα=β

Then

yeαt=∫γe(α−β)t dt+C=γα−βe(α−β)t+C.ye^{\alpha t}=\int \gamma e^{(\alpha-\beta)t}\,dt + C =\frac{\gamma}{\alpha-\beta}e^{(\alpha-\beta)t}+C.yeαt=∫γe(α−β)tdt+C=α−βγ​e(α−β)t+C.

Hence,

y(t)=γα−βe−βt+Ce−αt.y(t)=\frac{\gamma}{\alpha-\beta}e^{-\beta t}+Ce^{-\alpha t}.y(t)=α−βγ​e−βt+Ce−αt.

Now as t→∞t\to\inftyt→∞,

  • e−βt→0e^{-\beta t}\to 0e−βt→0 since β>0\beta>0β>0,
  • e−αt→0e^{-\alpha t}\to 0e−αt→0 since α>0\alpha>0α>0.

Therefore,

lim⁡t→∞y(t)=0.\lim_{t\to\infty} y(t)=0.t→∞lim​y(t)=0.

Case 2: α=β\alpha=\betaα=β

Then

ddt(yeαt)=γ.\frac{d}{dt}(ye^{\alpha t})=\gamma.dtd​(yeαt)=γ.

Integrating,

yeαt=γt+C.ye^{\alpha t}=\gamma t + C.yeαt=γt+C.

So,

y(t)=(γt+C)e−αt.y(t)=(\gamma t + C)e^{-\alpha t}.y(t)=(γt+C)e−αt.

Since α>0\alpha>0α>0,

lim⁡t→∞(γt+C)e−αt=0.\lim_{t\to\infty}(\gamma t + C)e^{-\alpha t}=0.t→∞lim​(γt+C)e−αt=0.

Thus again,

lim⁡t→∞y(t)=0.\lim_{t\to\infty} y(t)=0.t→∞lim​y(t)=0.
  1. Evaluate the options.
  • A: is 0 — Correct
  • B: is 1 — Incorrect
  • C: is −1-1−1 — Incorrect
  • D: does not exist — Incorrect

Therefore, the correct option is

A\boxed{\text{A}}A​

with value

0.\boxed{0}.0​.
  1. Comparison with stored correct answer:

Stored correct answer = A.

Our derived answer is also A, so they agree.

PreviousNext

More from Differential Equations

  • Let y=f(x) be the solution of the differential equation y(x+1)dx−x2dy=0,y(1)=e. Then x→0+lim​f(x) is equal to2023 · MCQ
  • Let y=y(x) be the solution of the differential equation xloge​xdxdy​+y=x2loge​x,(x>1). If y(2)=2, then y(e) is equal to2023 · MCQ
  • Let the solution curve y=y(x) of the differential equation  dxdy​−(1+x6)3/23x5tan−1(x3)​y=2xexp{(1+x6)​x3−tan−1x3​} pass through the origin. Then y(1) is equal to : …2023 · MCQ
  • The solution of the differential equation dxdy​=−(3x2+y2x2+3y2​),y(1)=0 is :2023 · MCQ
  • Let y=y(x) be the solution of the differential equation (3y2−5x2)y dx+2x(x2−y2)dy=0 such that y(1)=1. Then ​(y(2))3−12y(2)​ is equal to :2023 · MCQ
  • If x = x(y) is the solution of the differential equation ydydx​=2x+y3(y+1)ey,x(1)=0; then x(e) is equal to :2022 · MCQ
  • The slope of the tangent to a curve C:y=y(x) at any point (x,y) on it is 2+9e−2x2e2x−6e−x+9​. If C passes through the points $$\left(0, \frac{1}{2}+\frac{\pi}{2…2022 · MCQ
  • The general solution of the differential equation (x−y2)dx+y(5x+y2)dy=0 is :2022 · MCQ