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Differential Equations question

2022 · 26 Jul · Shift 2 · Q37
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  5. /2022 · 26 Jul · Shift 2 · Q37

Differential Equations question

2022 · 26 Jul · Shift 2 · Q37

JEE MainMathematicsDifferential EquationsNumerical+4 / −1
Suppose y=y(x)y=y(x)y=y(x) be the solution curve to the differential equation dydx−y=2−e−x\frac{d y}{d x}-y=2-e^{-x}dxdy​−y=2−e−x such that lim⁡x→∞y(x)\lim\limits_{x \rightarrow \infty} y(x)x→∞lim​y(x) is finite. If aaa and bbb are respectively the xxx- and yyy-intercepts of the tangent to the curve at x=0x=0x=0, then the value of a−4ba-4 ba−4b is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 3

  1. Solve the differential equation

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

This is a linear differential equation: dydx+P(x)y=Q(x),with P(x)=−1.\frac{dy}{dx}+P(x)y=Q(x), \quad \text{with } P(x)=-1.dxdy​+P(x)y=Q(x),with P(x)=−1.

The integrating factor is I.F.=e∫−1 dx=e−x.I.F.=e^{\int -1\,dx}=e^{-x}.I.F.=e∫−1dx=e−x.

Multiplying the equation by e−xe^{-x}e−x: e−xdydx−e−xy=2e−x−e−2x.e^{-x}\frac{dy}{dx}-e^{-x}y=2e^{-x}-e^{-2x}.e−xdxdy​−e−xy=2e−x−e−2x.

The left-hand side becomes ddx(ye−x)=2e−x−e−2x.\frac{d}{dx}(ye^{-x})=2e^{-x}-e^{-2x}.dxd​(ye−x)=2e−x−e−2x.

Integrating: ye−x=∫(2e−x−e−2x)dx+C.ye^{-x}=\int \left(2e^{-x}-e^{-2x}\right)dx+C.ye−x=∫(2e−x−e−2x)dx+C.

Now, ∫2e−xdx=−2e−x,\int 2e^{-x}dx=-2e^{-x},∫2e−xdx=−2e−x, ∫(−e−2x)dx=12e−2x.\int (-e^{-2x})dx=\frac{1}{2}e^{-2x}.∫(−e−2x)dx=21​e−2x.

So, ye−x=−2e−x+12e−2x+C.ye^{-x}=-2e^{-x}+\frac{1}{2}e^{-2x}+C.ye−x=−2e−x+21​e−2x+C.

Multiplying by exe^xex: y=−2+12e−x+Cex.y=-2+\frac{1}{2}e^{-x}+Ce^x.y=−2+21​e−x+Cex.

  1. Use the condition that lim⁡x→∞y(x)\lim_{x\to\infty}y(x)limx→∞​y(x) is finite

As x→∞x\to\inftyx→∞,

  • −2-2−2 remains finite,
  • 12e−x→0\frac12 e^{-x}\to 021​e−x→0,
  • CexCe^xCex is finite only if C=0C=0C=0.

Hence, C=0,C=0,C=0, and therefore y(x)=−2+12e−x.y(x)=-2+\frac12 e^{-x}.y(x)=−2+21​e−x.

  1. Find the point and slope at x=0x=0x=0

At x=0x=0x=0, y(0)=−2+12=−32.y(0)=-2+\frac12=-\frac32.y(0)=−2+21​=−23​.

Now from the differential equation, dydx=y+2−e−x.\frac{dy}{dx}=y+2-e^{-x}.dxdy​=y+2−e−x.

At x=0x=0x=0, y′(0)=y(0)+2−1=−32+1=−12.y'(0)=y(0)+2-1=-\frac32+1=-\frac12.y′(0)=y(0)+2−1=−23​+1=−21​.

So the tangent at x=0x=0x=0 passes through (0,−32)\left(0,-\frac32\right)(0,−23​) with slope m=−12.m=-\frac12.m=−21​.

  1. Equation of the tangent

Using point-slope form: y+32=−12(x−0).y+\frac32=-\frac12(x-0).y+23​=−21​(x−0). So, y=−12x−32.y=-\frac12x-\frac32.y=−21​x−23​.

  1. Find the intercepts

  • xxx-intercept: set y=0y=0y=0, 0=−12x−320=-\frac12x-\frac320=−21​x−23​ −12x=32-\frac12x=\frac32−21​x=23​ x=−3.x=-3.x=−3. Hence, a=−3.a=-3.a=−3.

  • yyy-intercept: set x=0x=0x=0, y=−32.y=-\frac32.y=−23​. Hence, b=−32.b=-\frac32.b=−23​.

  1. Compute a−4ba-4ba−4b

a−4b=−3−4(−32)=−3+6=3.a-4b=-3-4\left(-\frac32\right)=-3+6=3.a−4b=−3−4(−23​)=−3+6=3.

Therefore, the required integer is 3.\boxed{3}.3​.

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