WBJEE · Maths · Differentiation
The function \(y=e^{k x}\) satisfies \(\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=y \frac{d y}{d x}\). It is valid for
- A exactly one value of \(k\)
- B two distinct values of \(k\)
- C three distinct values of \(k\)
- D infinitely many values of \(k\)
Answer & Solution
Correct Answer
(C) three distinct values of \(k\)
Step-by-step Solution
Detailed explanation
Hint : \(y=e^{k x}\) \(\Rightarrow \frac{d y}{d x}=k e^{k x}=k y \Rightarrow \frac{d^2 y}{d x^2}=k^2 e^{k x}=k^2 y\) Now LHS \(=\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=\left(k^2 y+k y\right)(k y-y)=k(k+1)(k-1) y^2=k\left(k^2-1\right) y^2\)…
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