MHT CET · Maths · Differentiation
If \(y=2 \sin x+3 \cos x\) and \(y+A \frac{d^2 y}{d x^2}=B\), then the values of \(A\), \(\mathrm{B}\) are respectively
- A 0,1
- B 0,-1
- C -1,0
- D 1,0
Answer & Solution
Correct Answer
(D) 1,0
Step-by-step Solution
Detailed explanation
\(y=2 \sin x+3 \cos x \)
\( \therefore \frac{d y}{d x}=2 \cos x-3 \sin x \)
\( \therefore \frac{d^2 y}{d x^2}=-2 \sin x-3 \cos x=-(2 \sin x+3 \cos x)=-y \)
\( \therefore y+\frac{d^2 y}{d x^2}=0\)
We have \(y+A \frac{d^2 y}{d x^2}=B \Rightarrow A=1, B=0\)
\( \therefore \frac{d y}{d x}=2 \cos x-3 \sin x \)
\( \therefore \frac{d^2 y}{d x^2}=-2 \sin x-3 \cos x=-(2 \sin x+3 \cos x)=-y \)
\( \therefore y+\frac{d^2 y}{d x^2}=0\)
We have \(y+A \frac{d^2 y}{d x^2}=B \Rightarrow A=1, B=0\)
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