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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
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Verified Answer
The correct answer is:
three distinct values of \(k\)
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
\(=\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=\left(k^2 y+k y\right)(k y-y)=k(k+1)(k-1) y^2=k\left(k^2-1\right) y^2\)
Now RHS \(y \frac{d y}{d x}=k y^2, A / q~ k\left(k^2-1\right) y^2=k y^2 \Rightarrow k\left[k^2-2\right]=0 \Rightarrow k=0, k= \pm \sqrt{2}\)
\(\Rightarrow \frac{d y}{d x}=k e^{k x}=k y \Rightarrow \frac{d^2 y}{d x^2}=k^2 e^{k x}=k^2 y\)
Now LHS
\(=\left(\frac{d^2 y}{d x^2}+\frac{d y}{d x}\right)\left(\frac{d y}{d x}-y\right)=\left(k^2 y+k y\right)(k y-y)=k(k+1)(k-1) y^2=k\left(k^2-1\right) y^2\)
Now RHS \(y \frac{d y}{d x}=k y^2, A / q~ k\left(k^2-1\right) y^2=k y^2 \Rightarrow k\left[k^2-2\right]=0 \Rightarrow k=0, k= \pm \sqrt{2}\)
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