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UPSC 2018 Maths Optional Paper 1 Q6c — Step-by-Step Solution

13 marks · Section B

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

Question

Solve: $y''+16y=32\sec 2x$.

Technique

Variation of parameters; $u_1'=-y_2g/W$, $u_2'=y_1g/W$, $W=4$; simplify with double-angle identities and combine $\cos2x\cos4x+\sin2x\sin4x=\cos2x$.

Solution

Step 1 — Complementary function

Auxiliary equation $m^2+16=0\Rightarrow m=\pm4i$, so

Step 2 — Set up variation of parameters

The forcing $32\sec2x$ is not a UC (undetermined-coefficient) function, so use variation of parameters. With $y_1=\cos4x,\ y_2=\sin4x$ the Wronskian is

For $y_p=u_1y_1+u_2y_2$ with right-hand side $g(x)=32\sec2x$:

Step 3 — Compute $u_1$

Using $\sin4x=2\sin2x\cos2x$:

Step 4 — Compute $u_2$

Using $\cos4x=2\cos^2 2x-1$:

Integrate ($\int\sec2x\,dx=\tfrac12\ln|\sec2x+\tan2x|$):

Step 5 — Assemble the particular integral

Combine the elementary part: $8(\cos2x\cos4x+\sin2x\sin4x)=8\cos(4x-2x)=8\cos2x$. Hence

Step 6 — General solution

Answer