Q.4 (c) If y=acos(log x)+bsin(log x) , prove that x²y_(n+2)+(2n+1)xy_(n+1)+(n²+1)y_n=0 .

y₁=-a/xsin(log x)+b/xcos(log x)⇒ xy₁=-asin(log x)+bcos(log x) . Differentiate: xy₂+y₁=-1/x[acos(log x)+bsin(log x)]=-y/x⇒ x²y₂+xy₁+y=0 . Leibnitz's theorem (n times): Dⁿ(x²y₂)=x²y_(n+2)+n(2x)y_(n+1)+fracn(n-1)2(2)y_n ; Dⁿ(xy₁)=xy_(n+1)+ny_n ; Dⁿ(y)=y_n . Adding: x²y_(n+2)+(2n+1)xy_(n+1)+(n²-n+n+1)y_n=0 x²y_(n+2)+(2n+1)xy_(n+1)+(n²+1)y_n=0 (proved).

If \(y=a\cos(\log x)+b\sin(\log x)\), prove that \(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2+1)y_n=0\).

Q.4 (c) If \(y=a\cos(\log x)+b\sin(\log x)\), prove that \(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2+1)y_n=0\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

1 answer

Best answer

\(y_1=-\frac ax\sin(\log x)+\frac bx\cos(\log x)\Rightarrow xy_1=-a\sin(\log x)+b\cos(\log x)\).

Differentiate: \(xy_2+y_1=-\frac1x\left[a\cos(\log x)+b\sin(\log x)\right]=-\frac yx\Rightarrow x^2y_2+xy_1+y=0\).

Leibnitz's theorem (n times):

\(D^n(x^2y_2)=x^2y_{n+2}+n(2x)y_{n+1}+\frac{n(n-1)}2(2)y_n\); \(D^n(xy_1)=xy_{n+1}+ny_n\); \(D^n(y)=y_n\).

Adding: \(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2-n+n+1)y_n=0\)

\(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2+1)y_n=0\) (proved).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?

If \(y=a\cos(\log x)+b\sin(\log x)\), prove that \(x^2y_{n+2}+(2n+1)xy_{n+1}+(n^2+1)y_n=0\).

1 answer

Your answer