CUET Preparation Today
If $y=3\, cos (log x)+4 sin (log x),$ then choose correct option : |
$x^2y"-xy'+y=0$ $x^2y"+xy'0y=0$ $x^2y"+xy'+y=0$ $x^2y"-xy'+y=0$ |
$x^2y"+xy'+y=0$ |
The correct answer is Option (3) → $x^2y"+xy'+y=0$ $y=3\cos \log x+4 \sin \log x$ $y'=\frac{-3\sin\log x+4\cos\log x}{x}$ $y''=\frac{1(-3\sin\log x+4\cos\log x)}{x^2}-\frac{(3\cos\log x+4\sin\log x)}{x^2}$ $x^2y''=-xy'-y$ $x^2y''+xy'+y=0$ |
