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Differential Equations question

2020 · 7 Jan · Shift 2 · Q37
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  5. /2020 · 7 Jan · Shift 2 · Q37

Differential Equations question

2020 · 7 Jan · Shift 2 · Q37

JEE MainMathematicsDifferential EquationsMCQ+4 / −1
Let y = y(x) be the solution curve of the differential equation, (y2−x)dydx=1\left( {{y^2} - x} \right){{dy} \over {dx}} = 1(y2−x)dxdy​=1, satisfying y(0) = 1. This curve intersects the x-axis at a point whose abscissa is :
  1. A
    2 + e
  2. B
    -e
  3. C
    2
  4. D
    2 - e
View written solutionFree

Correct answer: D

  1. Given differential equation
(y2−x)dydx=1(y^2-x)\frac{dy}{dx}=1(y2−x)dxdy​=1

with initial condition

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

We need the point where the curve meets the xxx-axis, i.e. where

y=0.y=0.y=0.
  1. Rewrite the differential equation

From

(y2−x)dydx=1,(y^2-x)\frac{dy}{dx}=1,(y2−x)dxdy​=1,

we get

dydx=1y2−x.\frac{dy}{dx}=\frac{1}{y^2-x}.dxdy​=y2−x1​.

It is more convenient to write xxx as a function of yyy:

dxdy=y2−x.\frac{dx}{dy}=y^2-x.dydx​=y2−x.

So the equation becomes

dxdy+x=y2.\frac{dx}{dy}+x=y^2.dydx​+x=y2.

This is a linear differential equation in xxx with independent variable yyy.

  1. Solve the linear differential equation

The integrating factor is

IF=e∫1 dy=ey.IF=e^{\int 1\,dy}=e^y.IF=e∫1dy=ey.

Multiplying throughout by eye^yey:

eydxdy+eyx=y2ey.e^y\frac{dx}{dy}+e^y x=y^2 e^y.eydydx​+eyx=y2ey.

Thus,

ddy(xey)=y2ey.\frac{d}{dy}(x e^y)=y^2 e^y.dyd​(xey)=y2ey.

Integrate both sides:

xey=∫y2ey dy+C.x e^y=\int y^2 e^y\,dy + C.xey=∫y2eydy+C.

Now,

∫y2ey dy=ey(y2−2y+2).\int y^2 e^y\,dy=e^y(y^2-2y+2).∫y2eydy=ey(y2−2y+2).

Hence,

xey=ey(y2−2y+2)+C.x e^y=e^y(y^2-2y+2)+C.xey=ey(y2−2y+2)+C.

So,

x=y2−2y+2+Ce−y.x=y^2-2y+2+Ce^{-y}.x=y2−2y+2+Ce−y.
  1. Use the initial condition

Given y(0)=1y(0)=1y(0)=1, this means when x=0x=0x=0, we have y=1y=1y=1.

Substitute (x,y)=(0,1)(x,y)=(0,1)(x,y)=(0,1):

0=12−2(1)+2+Ce−1.0=1^2-2(1)+2+Ce^{-1}.0=12−2(1)+2+Ce−1.

That is,

0=1−2+2+Ce=1+Ce.0=1-2+2+\frac{C}{e}=1+\frac{C}{e}.0=1−2+2+eC​=1+eC​.

So,

C=−e.C=-e.C=−e.

Therefore the solution is

x=y2−2y+2−e1−y.x=y^2-2y+2-e^{1-y}.x=y2−2y+2−e1−y.
  1. Find where the curve meets the xxx-axis

On the xxx-axis,

y=0.y=0.y=0.

Substitute y=0y=0y=0 into the solution:

x=02−2(0)+2−e1−0=2−e.x=0^2-2(0)+2-e^{1-0}=2-e.x=02−2(0)+2−e1−0=2−e.

So the abscissa is

2−e.\boxed{2-e}.2−e​.
  1. Check with options

Option D is

2−e,2-e,2−e,

which matches our result.

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