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

2024 · 27 Jan · Shift 2 · Q42
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. /2024 · 27 Jan · Shift 2 · Q42

Differential Equations question

2024 · 27 Jan · Shift 2 · Q42

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
If y=y(x)y=y(x)y=y(x) is the solution curve of the differential equation (x2−4)dy−(y2−3y)dx=0,x>2,y(4)=32\left(x^2-4\right) \mathrm{d} y-\left(y^2-3 y\right) \mathrm{d} x=0, x\gt 2, y(4)=\frac{3}{2}(x2−4)dy−(y2−3y)dx=0,x>2,y(4)=23​ and the slope of the curve is never zero, then the value of y(10)y(10)y(10) equals :
  1. A
    31+(8)1/4\frac{3}{1+(8)^{1 / 4}}1+(8)1/43​
  2. B
    31−(8)1/4\frac{3}{1-(8)^{1 / 4}}1−(8)1/43​
  3. C
    31−22\frac{3}{1-2 \sqrt{2}}1−22​3​
  4. D
    31+22\frac{3}{1+2 \sqrt{2}}1+22​3​
View written solutionFree

Correct answer: A

  1. Write the differential equation in separable form

Given

(x2−4) dy−(y2−3y) dx=0(x^2-4)\,dy-(y^2-3y)\,dx=0(x2−4)dy−(y2−3y)dx=0

so

(x2−4)dydx=y2−3y=y(y−3).(x^2-4)\frac{dy}{dx}=y^2-3y=y(y-3).(x2−4)dxdy​=y2−3y=y(y−3).

Hence,

dyy(y−3)=dxx2−4.\frac{dy}{y(y-3)}=\frac{dx}{x^2-4}.y(y−3)dy​=x2−4dx​.
  1. Separate and integrate

We use partial fractions.

For the left side:

1y(y−3)=Ay+By−3.\frac{1}{y(y-3)}=\frac{A}{y}+\frac{B}{y-3}.y(y−3)1​=yA​+y−3B​.

So

1=A(y−3)+By.1=A(y-3)+By.1=A(y−3)+By.

Comparing coefficients,

A+B=0,−3A=1A+B=0,\qquad -3A=1A+B=0,−3A=1

which gives

A=−13,B=13.A=-\frac13,\qquad B=\frac13.A=−31​,B=31​.

Thus,

∫dyy(y−3)=13∫(1y−3−1y)dy=13ln⁡∣y−3y∣.\int \frac{dy}{y(y-3)}=\frac13\int\left(\frac{1}{y-3}-\frac{1}{y}\right)dy =\frac13\ln\left|\frac{y-3}{y}\right|.∫y(y−3)dy​=31​∫(y−31​−y1​)dy=31​ln​yy−3​​.

For the right side:

1x2−4=1(x−2)(x+2)=14(1x−2−1x+2).\frac{1}{x^2-4}=\frac{1}{(x-2)(x+2)} =\frac14\left(\frac{1}{x-2}-\frac{1}{x+2}\right).x2−41​=(x−2)(x+2)1​=41​(x−21​−x+21​).

Hence,

∫dxx2−4=14ln⁡∣x−2x+2∣.\int \frac{dx}{x^2-4} =\frac14\ln\left|\frac{x-2}{x+2}\right|.∫x2−4dx​=41​ln​x+2x−2​​.

Therefore,

13ln⁡∣y−3y∣=14ln⁡∣x−2x+2∣+C.\frac13\ln\left|\frac{y-3}{y}\right|=\frac14\ln\left|\frac{x-2}{x+2}\right|+C.31​ln​yy−3​​=41​ln​x+2x−2​​+C.

Multiply by 12:

4ln⁡∣y−3y∣=3ln⁡∣x−2x+2∣+C1.4\ln\left|\frac{y-3}{y}\right|=3\ln\left|\frac{x-2}{x+2}\right|+C_1.4ln​yy−3​​=3ln​x+2x−2​​+C1​.

So,

∣y−3y∣4=C∣x−2x+2∣3.\left|\frac{y-3}{y}\right|^4=C\left|\frac{x-2}{x+2}\right|^3.​yy−3​​4=C​x+2x−2​​3.

Equivalently,

y−3y=K(x−2x+2)3/4\frac{y-3}{y}=K\left(\frac{x-2}{x+2}\right)^{3/4}yy−3​=K(x+2x−2​)3/4

for some constant KKK.

  1. Use the initial condition

Given x>2x>2x>2, so x−2x+2>0\dfrac{x-2}{x+2}>0x+2x−2​>0. Also,

y(4)=32.y(4)=\frac32.y(4)=23​.

Then

y−3y=32−332=−1.\frac{y-3}{y}=\frac{\frac32-3}{\frac32}=-1.yy−3​=23​23​−3​=−1.

At x=4x=4x=4,

(x−2x+2)3/4=(26)3/4=(13)3/4.\left(\frac{x-2}{x+2}\right)^{3/4}=\left(\frac{2}{6}\right)^{3/4}=\left(\frac13\right)^{3/4}.(x+2x−2​)3/4=(62​)3/4=(31​)3/4.

Therefore,

−1=K(13)3/4⇒K=−33/4.-1=K\left(\frac13\right)^{3/4} \quad\Rightarrow\quad K=-3^{3/4}.−1=K(31​)3/4⇒K=−33/4.

Hence,

y−3y=−33/4(x−2x+2)3/4.\frac{y-3}{y}=-3^{3/4}\left(\frac{x-2}{x+2}\right)^{3/4}.yy−3​=−33/4(x+2x−2​)3/4.

This can be written as

1−3y=−33/4(x−2x+2)3/41-\frac{3}{y}=-3^{3/4}\left(\frac{x-2}{x+2}\right)^{3/4}1−y3​=−33/4(x+2x−2​)3/4

so

3y=1+33/4(x−2x+2)3/4.\frac{3}{y}=1+3^{3/4}\left(\frac{x-2}{x+2}\right)^{3/4}.y3​=1+33/4(x+2x−2​)3/4.

Thus,

y=31+33/4(x−2x+2)3/4.y=\frac{3}{1+3^{3/4}\left(\frac{x-2}{x+2}\right)^{3/4}}.y=1+33/4(x+2x−2​)3/43​.
  1. Find y(10)y(10)y(10)

At x=10x=10x=10,

x−2x+2=812=23.\frac{x-2}{x+2}=\frac{8}{12}=\frac23.x+2x−2​=128​=32​.

So,

33/4(23)3/4=(2)−3/4⋅33/4⋅2?3^{3/4}\left(\frac23\right)^{3/4}=(2)^{-3/4}\cdot 3^{3/4}\cdot 2^{?}33/4(32​)3/4=(2)−3/4⋅33/4⋅2?

More directly,

33/4(23)3/4=(3⋅23)3/4=23/4=81/4.3^{3/4}\left(\frac23\right)^{3/4}= \left(3\cdot\frac23\right)^{3/4}=2^{3/4}=8^{1/4}.33/4(32​)3/4=(3⋅32​)3/4=23/4=81/4.

Hence,

y(10)=31+81/4.y(10)=\frac{3}{1+8^{1/4}}.y(10)=1+81/43​.
  1. Check the condition that slope is never zero

Slope is

dydx=y(y−3)x2−4.\frac{dy}{dx}=\frac{y(y-3)}{x^2-4}.dxdy​=x2−4y(y−3)​.

For x>2x>2x>2, denominator is positive. If slope were zero, then y=0y=0y=0 or y=3y=3y=3. From the obtained solution with y(4)=32y(4)=\frac32y(4)=23​, we have 0<y<30<y<30<y<3, so neither happens. Thus the condition is satisfied.

  1. Compare with options
y(10)=31+81/4y(10)=\frac{3}{1+8^{1/4}}y(10)=1+81/43​

which is Option A.

PreviousNext

More from Differential Equations

  • If the solution curve, of the differential equation  dxdy​=x−yx+y−2​ passing through the point (2,1) is tan−1(x−1y−1​)−β1​loge​(α+(x−1y−1​)2)=loge​∣x−1∣…2024 · Numerical
  • A function y=f(x) satisfies f(x)sin2x+sinx−(1+cos2x)f′(x)=0 with condition f(0)=0. Then, f(2π​) is equal to2024 · MCQ
  • If the solution curve y=y(x) of the differential equation (1+y2)(1+loge​x)dx+xdy=0,x>0 passes through the point (1,1) and y(e)=β+tan(23​)α−tan(23​)​…2024 · Numerical
  • If sin(xy​)=loge​∣x∣+2α​ is the solution of the differential equation xcos(xy​)dxdy​=ycos(xy​)+x and y(1)=3π​, then α2…2024 · MCQ
  • Let y=y(x) be the solution of the differential equation secx dy+{2(1−x)tanx+x(2−x)}dx=0 such that y(0)=2. Then y(2) is equal to:2024 · MCQ
  • Let y=y(x) be the solution of the differential equation (1−x2)dy=[xy+(x3+2)3(1−x2)​]dx,−1<x<1,y(0)=0. If y(21​)=nm​,m…2024 · Numerical
  • Let Y=Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y−y=Y′(x)(X−x) and the co-ordinate axes, where (x,y) is any point on the curve, is always 2Y′(x)−y2​+1,Y′(x)eq0…2024 · Numerical
  • Let y=y(x) be the solution of the differential equation dxdy​=sinx(secx−sinxtanx)(tanx)+y​,x∈(0,2π​) satisfying the condition y(4π​)=2. Then, y(3π​)…2024 · MCQ