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Need Solution for R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise Revision Exercise Question 11 Maths Textbook Solution.

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Answer: xy^{11}+2y^{1}-xy+x^{2}-2=0

Hint: Find first and second derivative of equation put in given differential equation to verify.

Given: x y=a e^{x}+b e^{-x}+x^{2}, \quad x \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-x y+x^{2}-2 x

Solution:x y=a e^{x}+b e^{-x}+x^{2},

\mathrm{y}+\mathrm{x} \frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}=\mathrm{ae}^{\mathrm{x}}-\mathrm{be}^{-\mathrm{x}}+2 \mathrm{x} \ldots \text { differentiating using product rule }

\frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{x} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} \mathrm{x}^{2}}+\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{ae}^{\mathrm{x}}+\mathrm{be}^{-\mathrm{x}}+2 \ldots \text { differentiating again } w . r \cdot t \text { to } x \\

\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-2=\mathrm{ae}^{\mathrm{x}}+\mathrm{be}^{-\mathrm{x}} \\                            

\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-2=\mathrm{x} \mathrm{y}-\mathrm{x}^{2} \\                                                \left[\begin{array}{l} x y=a e^{x}+b e^{-x}+x^{2} \\ x y-x^{2}=a e^{x}+b e^{-x} \end{array}\right]

\mathrm{xy}^{{}''}+2 \mathrm{y}^{{}'}-\mathrm{x} \mathrm{y}+\mathrm{x}^{2}-2=0

Hence proved

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