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Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 15 Maths Textbook Solution.

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Answer: \frac{d^{3} y}{d x^{8}}-7\left(\frac{d y}{d x}\right)+6 y=0

Hint: The number constants is equals to the number time we differentiate.

Given:  y=ae^{2x}+be^{-3x}+ce^{x}

Solution: y=ae^{2x}+be^{-3x}+ce^{x}

\begin{aligned} &\frac{d y}{d x}=2 a e^{2 x}-3 b e^{-3 x}+c e^{x} \\ &\frac{d^{2} y}{d x^{2}}=4 a e^{2 x}+9 b e^{-3 x}+c e^{x} \\ &\frac{d^{3} y}{d x^{3}}=8 a e^{2 x}-27 b e^{-3 x}+c e^{x} \\ &\frac{d^{3} y}{d x^{3}}=7\left(2 a e^{2 x}-3 b e^{-3 x}+c e^{x}\right)-6\left(a e^{2 x}+b e^{-3 x}+c e^{x}\right) \\ &\frac{d^{3} y}{d x^{3}}=7\left(\frac{d y}{d x}\right)-6 y \\ &\frac{d^{3} y}{d x^{3}}-7\left(\frac{d y}{d x}\right)+6 y=0 \end{aligned}

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