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Differential Equations question

2022 · 29 Jul · Shift 2 · Q30
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  5. /2022 · 29 Jul · Shift 2 · Q30

Differential Equations question

2022 · 29 Jul · Shift 2 · Q30

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y=y(x)y=y(x)y=y(x) be the solution curve of the differential equation dydx+(2x2+11x+13x3+6x2+11x+6)y=(x+3)x+1,x>−1\frac{d y}{d x}+\left(\frac{2 x^{2}+11 x+13}{x^{3}+6 x^{2}+11 x+6}\right) y=\frac{(x+3)}{x+1}, x\gt -1dxdy​+(x3+6x2+11x+62x2+11x+13​)y=x+1(x+3)​,x>−1, which passes through the point (0,1)(0,1)(0,1). Then y(1)y(1)y(1) is equal to :
  1. A
    12\frac{1}{2}21​
  2. B
    32\frac{3}{2}23​
  3. C
    52\frac{5}{2}25​
  4. D
    72\frac{7}{2}27​
View written solutionFree

Correct answer: B

  1. Write the differential equation in linear form

We have

dydx+(2x2+11x+13x3+6x2+11x+6)y=x+3x+1,x>−1\frac{dy}{dx}+\left(\frac{2x^2+11x+13}{x^3+6x^2+11x+6}\right)y=\frac{x+3}{x+1}, \qquad x>-1dxdy​+(x3+6x2+11x+62x2+11x+13​)y=x+1x+3​,x>−1

with initial condition

y(0)=1.y(0)=1.y(0)=1.

This is a first-order linear differential equation:

dydx+P(x)y=Q(x),\frac{dy}{dx}+P(x)y=Q(x),dxdy​+P(x)y=Q(x),

where

P(x)=2x2+11x+13x3+6x2+11x+6,Q(x)=x+3x+1.P(x)=\frac{2x^2+11x+13}{x^3+6x^2+11x+6}, \qquad Q(x)=\frac{x+3}{x+1}.P(x)=x3+6x2+11x+62x2+11x+13​,Q(x)=x+1x+3​.
  1. Factor the denominator of P(x)P(x)P(x)

Observe that

x3+6x2+11x+6=(x+1)(x+2)(x+3).x^3+6x^2+11x+6=(x+1)(x+2)(x+3).x3+6x2+11x+6=(x+1)(x+2)(x+3).

So

P(x)=2x2+11x+13(x+1)(x+2)(x+3).P(x)=\frac{2x^2+11x+13}{(x+1)(x+2)(x+3)}.P(x)=(x+1)(x+2)(x+3)2x2+11x+13​.

Now notice

ddx[(x+1)(x+2)(x+3)]=3x2+12x+11,\frac{d}{dx}\big[(x+1)(x+2)(x+3)\big]=3x^2+12x+11,dxd​[(x+1)(x+2)(x+3)]=3x2+12x+11,

which is not the numerator. So we use partial fractions.

Let

2x2+11x+13(x+1)(x+2)(x+3)=Ax+1+Bx+2+Cx+3.\frac{2x^2+11x+13}{(x+1)(x+2)(x+3)}=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{x+3}.(x+1)(x+2)(x+3)2x2+11x+13​=x+1A​+x+2B​+x+3C​.

Then

2x2+11x+13=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2).2x^2+11x+13=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2).2x2+11x+13=A(x+2)(x+3)+B(x+1)(x+3)+C(x+1)(x+2).

Substitute convenient values:

  • At x=−1x=-1x=−1: 2−11+13=4=2A  ⟹  A=2.2-11+13=4=2A \implies A=2.2−11+13=4=2A⟹A=2.
  • At x=−2x=-2x=−2: 8−22+13=−1=−B  ⟹  B=1.8-22+13=-1=-B \implies B=1.8−22+13=−1=−B⟹B=1.
  • At x=−3x=-3x=−3: 18−33+13=−2=2C  ⟹  C=−1.18-33+13=-2=2C \implies C=-1.18−33+13=−2=2C⟹C=−1.

Hence

P(x)=2x+1+1x+2−1x+3.P(x)=\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}.P(x)=x+12​+x+21​−x+31​.
  1. Find the integrating factor

The integrating factor is

I.F.=e∫P(x) dx=e∫(2x+1+1x+2−1x+3)dx.I.F.=e^{\int P(x)\,dx} =e^{\int \left(\frac{2}{x+1}+\frac{1}{x+2}-\frac{1}{x+3}\right)dx}.I.F.=e∫P(x)dx=e∫(x+12​+x+21​−x+31​)dx.

So

I.F.=e2ln⁡(x+1)+ln⁡(x+2)−ln⁡(x+3)=(x+1)2(x+2)x+3.I.F.=e^{2\ln(x+1)+\ln(x+2)-\ln(x+3)} =\frac{(x+1)^2(x+2)}{x+3}.I.F.=e2ln(x+1)+ln(x+2)−ln(x+3)=x+3(x+1)2(x+2)​.

Since x>−1x>-1x>−1, all factors are positive, so no absolute value issue arises.

  1. Multiply the equation by the integrating factor

Then

ddx[y⋅(x+1)2(x+2)x+3]=(x+1)2(x+2)x+3⋅x+3x+1.\frac{d}{dx}\left[y\cdot \frac{(x+1)^2(x+2)}{x+3}\right] =\frac{(x+1)^2(x+2)}{x+3}\cdot \frac{x+3}{x+1}.dxd​[y⋅x+3(x+1)2(x+2)​]=x+3(x+1)2(x+2)​⋅x+1x+3​.

The right-hand side simplifies to

(x+1)(x+2).(x+1)(x+2).(x+1)(x+2).

Therefore,

ddx[y⋅(x+1)2(x+2)x+3]=(x+1)(x+2).\frac{d}{dx}\left[y\cdot \frac{(x+1)^2(x+2)}{x+3}\right]=(x+1)(x+2).dxd​[y⋅x+3(x+1)2(x+2)​]=(x+1)(x+2).
  1. Integrate both sides

Integrating,

y⋅(x+1)2(x+2)x+3=∫(x+1)(x+2) dx+C.y\cdot \frac{(x+1)^2(x+2)}{x+3}=\int (x+1)(x+2)\,dx + C.y⋅x+3(x+1)2(x+2)​=∫(x+1)(x+2)dx+C.

Now

(x+1)(x+2)=x2+3x+2,(x+1)(x+2)=x^2+3x+2,(x+1)(x+2)=x2+3x+2,

so

∫(x2+3x+2)dx=x33+3x22+2x+C.\int (x^2+3x+2)dx=\frac{x^3}{3}+\frac{3x^2}{2}+2x+C.∫(x2+3x+2)dx=3x3​+23x2​+2x+C.

Thus

y⋅(x+1)2(x+2)x+3=x33+3x22+2x+C.y\cdot \frac{(x+1)^2(x+2)}{x+3}=\frac{x^3}{3}+\frac{3x^2}{2}+2x+C.y⋅x+3(x+1)2(x+2)​=3x3​+23x2​+2x+C.
  1. Use the initial condition y(0)=1y(0)=1y(0)=1

At x=0x=0x=0,

1⋅(1)2(2)3=C.1\cdot \frac{(1)^2(2)}{3}=C.1⋅3(1)2(2)​=C.

Hence

C=23.C=\frac{2}{3}.C=32​.

So

y⋅(x+1)2(x+2)x+3=x33+3x22+2x+23.y\cdot \frac{(x+1)^2(x+2)}{x+3}=\frac{x^3}{3}+\frac{3x^2}{2}+2x+\frac{2}{3}.y⋅x+3(x+1)2(x+2)​=3x3​+23x2​+2x+32​.
  1. Compute y(1)y(1)y(1)

Put x=1x=1x=1:

y(1)⋅(2)2(3)4=13+32+2+23.y(1)\cdot \frac{(2)^2(3)}{4}=\frac{1}{3}+\frac{3}{2}+2+\frac{2}{3}.y(1)⋅4(2)2(3)​=31​+23​+2+32​.

Left side:

(2)2(3)4=3,\frac{(2)^2(3)}{4}=3,4(2)2(3)​=3,

so left side is 3y(1)3y(1)3y(1).

Right side:

13+23=1,\frac{1}{3}+\frac{2}{3}=1,31​+32​=1,

therefore

1+32+2=92.1+\frac{3}{2}+2=\frac{9}{2}.1+23​+2=29​.

So

3y(1)=92  ⟹  y(1)=96=32.3y(1)=\frac{9}{2} \implies y(1)=\frac{9}{6}=\frac{3}{2}.3y(1)=29​⟹y(1)=69​=23​.
  1. Match with the options
y(1)=32y(1)=\frac{3}{2}y(1)=23​

So the correct option is:

B. 32\frac{3}{2}23​

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