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  • Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 11

Need solution for RD Sharma maths class 12 chapter 21 Differential Equation exercise Fill in the blank question 11

Answers (1)

Answer:

 1

Hint:

 Squaring both sides

Given:

 \sqrt{1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}}=\frac{\mathrm{d} y}{\mathrm{d} x}+x

Solution:

On squaring the given differential equation,

\begin{aligned} &\left ( \sqrt{1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}}\right )^{2} =\left ( \frac{\mathrm{d} y}{\mathrm{d} x}+x \right )^{2} \\ &\Rightarrow 1+\frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=\left ( \frac{\mathrm{d} y}{\mathrm{d} x}+x \right )^{2} \end{aligned}

Highest order of the given equation is

\begin{aligned} & \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}}=2 \end{aligned}

and degree of

\begin{aligned} & \frac{\mathrm{d} ^{2}y}{\mathrm{d} x^{2}} \, \, is\, \, 1 \end{aligned}

∴ Degree of the differential equation is 1.

∴So, the answer is 1.

Posted by

Gurleen Kaur

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