TS EAMCET · Maths · Differential Equations
If the family of curves \(y=a e^{4 x}+b e^{-x}\), where \(a, b\) are arbitrary constants represents the general solution of the differential equation \(f\left(x, y \frac{d y}{d x}, \frac{d^2 y}{d x^2}\right)=0\), then \(\frac{d f}{d x}=\)
- A \(\frac{d^2 y}{d x^2}-3 \frac{d y}{d x}-4 y\)
- B \(\frac{d^3 y}{d x^3}-3 \frac{d^2 y}{d x^2}-4 \frac{d y}{d x}\)
- C \(\frac{d^3 y}{d x^3}-\frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+2\)
- D \(\frac{d^3 y}{d x^3}-\frac{d^2 y}{d x^2}+3\)
Answer & Solution
Correct Answer
(B) \(\frac{d^3 y}{d x^3}-3 \frac{d^2 y}{d x^2}-4 \frac{d y}{d x}\)
Step-by-step Solution
Detailed explanation
We have, \(y=a e^{4 x}+b e^{-x}\) ...(i) \(\therefore \quad \frac{d y}{d x}=4 a e^{4 x}-b e^{-x}\) ...(ii) and \(\frac{d^2 y}{d x^2}=16 a e^{4 x}+b e^{-x}\) ...(iii) On adding Eqs. (i) and (ii), we get \(y+\frac{d y}{d x}=5 a e^{4 x}\)…
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