← 2017 Paper 1

UPSC 2017 Maths Optional Paper 1 Q6a-i — Step-by-Step Solution

8 marks · Section B

Particular integral via operator method · ODEs · asked 7× in 13 yrs · Read the full method →

Question

Solve the following simultaneous linear differential equations: (D+1)y=z+ex(D+1)y=z+e^x and (D+1)z=y+ex(D+1)z=y+e^x where yy and zz are functions of independent variable xx and DddxD\equiv\dfrac{d}{dx}.

Technique

Symmetric system — decouple by forming y+zy+z and yzy-z, solve each first-order equation, recombine.

Solution

Step 1 — Add and subtract to decouple

Let u=y+zu=y+z and w=yzw=y-z. Adding the two equations:

(D+1)(y+z)=(y+z)+2ex  (D+1)u=u+2ex  Du=2ex.(D+1)(y+z)=(y+z)+2e^x\ \Rightarrow\ (D+1)u=u+2e^x\ \Rightarrow\ Du=2e^x.

Subtracting (first minus second):

(D+1)(yz)=(zy)+0  (D+1)w=w  (D+2)w=0.(D+1)(y-z)=(z-y)+0\ \Rightarrow\ (D+1)w=-w\ \Rightarrow\ (D+2)w=0.

Step 2 — Solve the two decoupled equations

For uu: dudx=2exu=2ex+2C1\dfrac{du}{dx}=2e^x\Rightarrow u=2e^x+2C_1 (write the constant as 2C12C_1 for convenience).

For ww: dwdx=2ww=2C2e2x\dfrac{dw}{dx}=-2w\Rightarrow w=2C_2e^{-2x}.

Step 3 — Recover yy and zz

y=u+w2=C1+ex+C2e2x,z=uw2=C1+exC2e2x.y=\frac{u+w}{2}=C_1+e^x+C_2e^{-2x},\qquad z=\frac{u-w}{2}=C_1+e^x-C_2e^{-2x}.

Answer

  y=C1+C2e2x+ex,z=C1C2e2x+ex.  \boxed{\;y=C_1+C_2e^{-2x}+e^x,\qquad z=C_1-C_2e^{-2x}+e^x.\;}
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