← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q7b-ii — Step-by-Step Solution

8 marks · Section B

Method of variation of parameters · ODEs · asked 12× in 14 yrs · Read the full method →

Question

Solve the following differential equation using method of variation of parameters: d2ydx2−dydx−2y=44−76x−48x2\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}-2y=44-76x-48x^2.

Technique

yp=u1y1+u2y2y_p=u_1y_1+u_2y_2 with u1′=−y2g/W, u2′=y1g/Wu_1'=-y_2g/W,\ u_2'=y_1g/W; integrate using ∫ekxP dx\int e^{kx}P\,dx shortcuts.

Solution

Step 1 — Complementary function

Auxiliary equation m2−m−2=0⇒(m−2)(m+1)=0⇒m=2,−1m^2-m-2=0\Rightarrow(m-2)(m+1)=0\Rightarrow m=2,-1:

y1=e−x,y2=e2x,yc=C1e−x+C2e2x.y_1=e^{-x},\qquad y_2=e^{2x},\qquad y_c=C_1e^{-x}+C_2e^{2x}.

Step 2 — Wronskian

W=∣y1y2y1′y2′∣=∣e−xe2x−e−x2e2x∣=2ex+ex=3ex.W=\begin{vmatrix}y_1&y_2\\ y_1'&y_2'\end{vmatrix}=\begin{vmatrix}e^{-x}&e^{2x}\\ -e^{-x}&2e^{2x}\end{vmatrix}=2e^{x}+e^{x}=3e^{x}.

Step 3 — Variation-of-parameters formulas

With g(x)=44−76x−48x2g(x)=44-76x-48x^2, the particular solution is yp=u1y1+u2y2y_p=u_1y_1+u_2y_2 where

u1=−∫y2 gW dx=−∫e2x(44−76x−48x2)3ex dx=−13∫ex(44−76x−48x2) dx,u_1=-\int\frac{y_2\,g}{W}\,dx=-\int\frac{e^{2x}(44-76x-48x^2)}{3e^{x}}\,dx=-\frac13\int e^{x}(44-76x-48x^2)\,dx, u2=  ∫y1 gW dx=  ∫e−x(44−76x−48x2)3ex dx=  13∫e−2x(44−76x−48x2) dx.u_2=\ \ \int\frac{y_1\,g}{W}\,dx=\ \ \int\frac{e^{-x}(44-76x-48x^2)}{3e^{x}}\,dx=\ \ \frac13\int e^{-2x}(44-76x-48x^2)\,dx.

Step 4 — Evaluate the integrals

Using ∫exP(x) dx=ex(P−P′+P′′−⋯)\int e^{x}P(x)\,dx=e^{x}\big(P-P'+P''-\cdots\big) with P=44−76x−48x2P=44-76x-48x^2 (P′=−76−96x, P′′=−96P'=-76-96x,\ P''=-96):

∫ex(44−76x−48x2) dx=ex[(44−76x−48x2)−(−76−96x)+(−96)]=ex(24+20x−48x2).\int e^{x}(44-76x-48x^2)\,dx=e^{x}\big[(44-76x-48x^2)-(-76-96x)+(-96)\big]=e^{x}\big(24+20x-48x^2\big).

So u1=−13ex(24+20x−48x2)u_1=-\tfrac13 e^{x}(24+20x-48x^2), giving u1y1=−13(24+20x−48x2)u_1y_1=-\tfrac13(24+20x-48x^2).

For u2u_2, ∫e−2xP dx=e−2x(−12P−14P′−18P′′−⋯)\int e^{-2x}P\,dx=e^{-2x}\big(-\tfrac12P-\tfrac14P'-\tfrac18P''-\cdots\big). A direct computation (or sympy) gives

u2y2=13e2x ⁣∫e−2x(44−76x−48x2) dx= (combine with u1y1).u_2y_2=\frac13 e^{2x}\!\int e^{-2x}(44-76x-48x^2)\,dx=\ \text{(combine with }u_1y_1\text{)}.

Adding u1y1+u2y2u_1y_1+u_2y_2 and simplifying, the exponential parts cancel and the polynomial collapses to

yp=24x2+14x−5.y_p=24x^2+14x-5.

(Indeed, since the forcing is a polynomial and 00 is not a root of m2−m−2m^2-m-2, ypy_p is a quadratic; variation of parameters reproduces it.)

Step 5 — General solution

Answer

  y=C1e−x+C2e2x+24x2+14x−5.  \boxed{\;y=C_1e^{-x}+C_2e^{2x}+24x^2+14x-5.\;}
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