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MathHardAdv 2021 • Paper 2

Solution to First Order Linear Differential Equation with Given Boundary Condition

For any real numbers α\alpha and β\beta, let yα,β(x)y_{\alpha,\beta}(x), x∈Rx \in \mathbb{R}, be the solution of the differential equation dydx+αy=xeβx,y(1)=1.\frac{dy}{dx} + \alpha y = x e^{\beta x}, \quad y(1) = 1. Let S={yα,β(x):α,β∈R}S = \{y_{\alpha,\beta}(x) : \alpha, \beta \in \mathbb{R}\}. Which of the given functions belong(s) to the set SS?

Options

A

f(x)=x22e−x+(e−12)e−xf(x) = \frac{x^2}{2} e^{-x} + \left(e - \frac{1}{2}\right) e^{-x}

Correct
B

f(x)=−x22e−x+(e+12)e−xf(x) = -\frac{x^2}{2} e^{-x} + \left(e + \frac{1}{2}\right) e^{-x}

C

f(x)=ex2(x−12)+(e−e24)e−xf(x) = \frac{e^x}{2} \left(x - \frac{1}{2}\right) + \left(e - \frac{e^2}{4}\right) e^{-x}

Correct
D

f(x)=ex2(12−x)+(e+e24)e−xf(x) = \frac{e^x}{2} \left(\frac{1}{2} - x\right) + \left(e + \frac{e^2}{4}\right) e^{-x}

Topics & Concepts

Differential EquationsDifferential EquationsLinear Differential Equations

Step-by-Step Solution

To determine which of the given functions belong to the set S={yα,β(x):α,β∈R}S = \{y_{\alpha,\beta}(x) : \alpha, \beta \in \mathbb{R}\}, we need to solve the given first-order linear differential equation:

dydx+αy=xeβx,with y(1)=1\frac{dy}{dx} + \alpha y = x e^{\beta x}, \quad \text{with } y(1) = 1

Step 1: Finding the Integrating Factor

The differential equation is of the standard linear form dydx+P(x)y=Q(x)\frac{dy}{dx} + P(x) y = Q(x), where P(x)=αP(x) = \alpha and Q(x)=xeβxQ(x) = x e^{\beta x}.

The Integrating Factor (I.F.) is: I.F.=e∫α dx=eαx\text{I.F.} = e^{\int \alpha \, dx} = e^{\alpha x}

Multiplying the differential equation by the integrating factor gives: eαxdydx+αeαxy=xe(α+β)xe^{\alpha x} \frac{dy}{dx} + \alpha e^{\alpha x} y = x e^{(\alpha + \beta) x} ddx(yeαx)=xe(α+β)x\frac{d}{dx} \left( y e^{\alpha x} \right) = x e^{(\alpha + \beta) x}

Integrating both sides with respect to xx: yeαx=∫xe(α+β)x dxy e^{\alpha x} = \int x e^{(\alpha + \beta) x} \, dx

Step 2: Case Analysis

Case 1: When α+β=0\alpha + \beta = 0 (i.e., β=−α\beta = -\alpha)

The equation simplifies to: ddx(yeαx)=x\frac{d}{dx} \left( y e^{\alpha x} \right) = x

Integrating both sides: yeαx=x22+Cy e^{\alpha x} = \frac{x^2}{2} + C y(x)=x22e−αx+Ce−αxy(x) = \frac{x^2}{2} e^{-\alpha x} + C e^{-\alpha x}

Using the initial condition y(1)=1y(1) = 1: 1=12e−α+Ce−α  ⟹  eα=12+C  ⟹  C=eα−121 = \frac{1}{2} e^{-\alpha} + C e^{-\alpha} \implies e^{\alpha} = \frac{1}{2} + C \implies C = e^{\alpha} - \frac{1}{2}

Thus, the solution for β=−α\beta = -\alpha is: yα,−α(x)=x22e−αx+(eα−12)e−αxy_{\alpha, -\alpha}(x) = \frac{x^2}{2} e^{-\alpha x} + \left( e^{\alpha} - \frac{1}{2} \right) e^{-\alpha x}

If we choose α=1\alpha = 1 and β=−1\beta = -1: y1,−1(x)=x22e−x+(e−12)e−xy_{1,-1}(x) = \frac{x^2}{2} e^{-x} + \left( e - \frac{1}{2} \right) e^{-x}

This matches Option (A).

Case 2: When α+β≠0\alpha + \beta \neq 0

Using integration by parts for ∫xe(α+β)x dx\int x e^{(\alpha + \beta) x} \, dx: ∫xe(α+β)x dx=x(e(α+β)xα+β)−∫e(α+β)xα+β dx=e(α+β)xα+β(x−1α+β)\int x e^{(\alpha + \beta) x} \, dx = x \left( \frac{e^{(\alpha + \beta) x}}{\alpha + \beta} \right) - \int \frac{e^{(\alpha + \beta) x}}{\alpha + \beta} \, dx = \frac{e^{(\alpha + \beta) x}}{\alpha + \beta} \left( x - \frac{1}{\alpha + \beta} \right)

So the general solution becomes: yeαx=e(α+β)xα+β(x−1α+β)+Cy e^{\alpha x} = \frac{e^{(\alpha + \beta) x}}{\alpha + \beta} \left( x - \frac{1}{\alpha + \beta} \right) + C y(x)=eβxα+β(x−1α+β)+Ce−αxy(x) = \frac{e^{\beta x}}{\alpha + \beta} \left( x - \frac{1}{\alpha + \beta} \right) + C e^{-\alpha x}

Using the initial condition y(1)=1y(1) = 1: 1=eβα+β(1−1α+β)+Ce−α1 = \frac{e^{\beta}}{\alpha + \beta} \left( 1 - \frac{1}{\alpha + \beta} \right) + C e^{-\alpha} C=eα−eα+βα+β(1−1α+β)C = e^{\alpha} - \frac{e^{\alpha + \beta}}{\alpha + \beta} \left( 1 - \frac{1}{\alpha + \beta} \right)

If we choose α=1\alpha = 1 and β=1\beta = 1 (so α+β=2\alpha + \beta = 2): C=e1−e22(1−12)=e−e24C = e^1 - \frac{e^2}{2} \left( 1 - \frac{1}{2} \right) = e - \frac{e^2}{4}

Substituting α=1\alpha = 1, β=1\beta = 1, and C=e−e24C = e - \frac{e^2}{4} back into y(x)y(x): y1,1(x)=ex2(x−12)+(e−e24)e−xy_{1,1}(x) = \frac{e^x}{2} \left( x - \frac{1}{2} \right) + \left( e - \frac{e^2}{4} \right) e^{-x}

This matches Option (C).

Conclusion

  • Option (A) corresponds to α=1,β=−1\alpha = 1, \beta = -1.
  • Option (C) corresponds to α=1,β=1\alpha = 1, \beta = 1.

Correct Options: (A) and (C)

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