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Differential Equations question

2022 · 28 Jun · Shift 1 · Q32
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  5. /2022 · 28 Jun · Shift 1 · Q32

Differential Equations question

2022 · 28 Jun · Shift 1 · Q32

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution of the differential equation x(1−x2)dydx+(3x2y−y−4x3)=0x(1 - {x^2}){{dy} \over {dx}} + (3{x^2}y - y - 4{x^3}) = 0x(1−x2)dxdy​+(3x2y−y−4x3)=0, x>1x \gt 1x>1, with y(2)=−2y(2) = - 2y(2)=−2. Then y(3) is equal to :
  1. A
    −-− 18
  2. B
    −-− 12
  3. C
    −-− 6
  4. D
    −-− 3
View written solutionFree

Correct answer: A

  1. Given differential equation

x(1−x2)dydx+(3x2y−y−4x3)=0,x(1-x^2)\frac{dy}{dx} + (3x^2y-y-4x^3)=0,x(1−x2)dxdy​+(3x2y−y−4x3)=0, with x>1x>1x>1 and y(2)=−2y(2)=-2y(2)=−2.

We need to find y(3)y(3)y(3).

  1. Rewrite the equation in linear form

First expand the middle term:

3x2y−y=y(3x2−1).3x^2y-y = y(3x^2-1).3x2y−y=y(3x2−1).

So the equation becomes

x(1−x2)dydx+(3x2−1)y−4x3=0.x(1-x^2)\frac{dy}{dx} + (3x^2-1)y - 4x^3 = 0.x(1−x2)dxdy​+(3x2−1)y−4x3=0.

Hence,

x(1−x2)dydx+(3x2−1)y=4x3.x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = 4x^3.x(1−x2)dxdy​+(3x2−1)y=4x3.

Now divide by x(1−x2)x(1-x^2)x(1−x2):

dydx+3x2−1x(1−x2)y=4x21−x2.\frac{dy}{dx} + \frac{3x^2-1}{x(1-x^2)}y = \frac{4x^2}{1-x^2}.dxdy​+x(1−x2)3x2−1​y=1−x24x2​.

  1. Observe a useful substitution

Notice that

ddx[y x (1−x2)]=x(1−x2)dydx+yddx[x(1−x2)].\frac{d}{dx}\big[y\,x\,(1-x^2)\big] = x(1-x^2)\frac{dy}{dx} + y\frac{d}{dx}[x(1-x^2)].dxd​[yx(1−x2)]=x(1−x2)dxdy​+ydxd​[x(1−x2)].

Now,

ddx[x(1−x2)]=ddx(x−x3)=1−3x2.\frac{d}{dx}[x(1-x^2)] = \frac{d}{dx}(x-x^3)=1-3x^2.dxd​[x(1−x2)]=dxd​(x−x3)=1−3x2.

Therefore,

ddx[yx(1−x2)]=x(1−x2)dydx+y(1−3x2).\frac{d}{dx}\big[yx(1-x^2)\big] = x(1-x^2)\frac{dy}{dx} + y(1-3x^2).dxd​[yx(1−x2)]=x(1−x2)dxdy​+y(1−3x2).

But in the given equation we have

x(1−x2)dydx+y(3x2−1).x(1-x^2)\frac{dy}{dx} + y(3x^2-1).x(1−x2)dxdy​+y(3x2−1).

So,

= x(1-x^2)\frac{dy}{dx} - y(1-3x^2) = \frac{d}{dx}[yx(1-x^2)]\text{ with a minus sign adjustment.}$$ More directly, $$x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = x(1-x^2)\frac{dy}{dx} - (1-3x^2)y.$$ Hence, $$x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = \frac{d}{dx}[yx(1-x^2)].$$ So the differential equation simplifies to $$\frac{d}{dx}[yx(1-x^2)] = 4x^3.$$ --- 4. **Integrate both sides** $$yx(1-x^2) = \int 4x^3\,dx = x^4 + C.$$ Thus, $$y = \frac{x^4+C}{x(1-x^2)}.$$ --- 5. **Use the initial condition $y(2)=-2$** Substitute $x=2$: $$-2 = \frac{2^4 + C}{2(1-2^2)} = \frac{16+C}{2(1-4)} = \frac{16+C}{-6}.$$ So, $$-2 = \frac{16+C}{-6}.$$ Multiply both sides by $-6$: $$12 = 16 + C.$$ Hence, $$C = -4.$$ Therefore, $$y = \frac{x^4-4}{x(1-x^2)}.$$ --- 6. **Find $y(3)$** $$y(3)=\frac{3^4-4}{3(1-3^2)} = \frac{81-4}{3(1-9)} = \frac{77}{3(-8)} = -\frac{77}{24}.$$ This does not match any option, so let us recheck the simplification. From the original equation: $$x(1-x^2)\frac{dy}{dx} + (3x^2y-y-4x^3)=0$$ $$\Rightarrow x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = 4x^3.$$ Now compute: $$\frac{d}{dx}[xy(1-x^2)] = x(1-x^2)\frac{dy}{dx} + y(1-3x^2).$$ Therefore, $$x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = x(1-x^2)\frac{dy}{dx} - (1-3x^2)y.$$ So actually, $$x(1-x^2)\frac{dy}{dx} + (3x^2-1)y \neq \frac{d}{dx}[xy(1-x^2)],$$ but $$x(1-x^2)\frac{dy}{dx} + (3x^2-1)y = x(1-x^2)\frac{dy}{dx} - y(1-3x^2).$$ This suggests using integrating factor method properly. --- 7. **Solve by integrating factor** We have $$\frac{dy}{dx} + \frac{3x^2-1}{x(1-x^2)}y = \frac{4x^2}{1-x^2}.$$ Now, $$\frac{3x^2-1}{x(1-x^2)} = -\frac{3x^2-1}{x(x^2-1)} = \frac{1-3x^2}{x(1-x^2)}.$$ But notice $$\frac{d}{dx}\ln[x(1-x^2)] = \frac{1-3x^2}{x(1-x^2)}.$$ Hence, $$P(x)=\frac{3x^2-1}{x(1-x^2)} = \frac{d}{dx}\ln\left(\frac{1}{x(1-x^2)}\right)$$ or more simply, $$P(x)= -\frac{d}{dx}\ln[x(1-x^2)].$$ Thus the integrating factor is $$\text{I.F.} = e^{\int P(x)dx} = e^{-\ln[x(1-x^2)]} = \frac{1}{x(1-x^2)}.$$ Multiply the equation by the integrating factor: $$\frac{1}{x(1-x^2)}\frac{dy}{dx} + \frac{3x^2-1}{x^2(1-x^2)^2}y = \frac{4x}{(1-x^2)^2}.$$ So, $$\frac{d}{dx}\left(\frac{y}{x(1-x^2)}\right)=\frac{4x}{(1-x^2)^2}.$$ Integrate: $$\frac{y}{x(1-x^2)} = \int \frac{4x}{(1-x^2)^2}dx.$$ Let $$u=1-x^2 \Rightarrow du=-2x\,dx,$$ so $$\int \frac{4x}{(1-x^2)^2}dx = \int \frac{4x}{u^2}dx = -2\int u^{-2}du = -2\left(\frac{u^{-1}}{-1}\right)=\frac{2}{u} = \frac{2}{1-x^2}.$$ Therefore, $$\frac{y}{x(1-x^2)} = \frac{2}{1-x^2} + C.$$ Multiply through: $$y = 2x + Cx(1-x^2).$$ --- 8. **Use the condition $y(2)=-2$** $$-2 = 2(2) + C\cdot 2(1-4)$$ $$-2 = 4 - 6C$$ $$-6 = -6C$$ $$C=1.$$ Thus, $$y = 2x + x(1-x^2) = 3x - x^3.$$ --- 9. **Compute $y(3)$** $$y(3)=3(3)-3^3=9-27=-18.$$ --- 10. **Check options** The correct option is: $$\boxed{\text{A }(-18)}$$
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