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What is the degree of the differential equation
$$
\frac{d^{2} y}{d x^{2}}-\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}=0 ?
$$
Options:
$$
\frac{d^{2} y}{d x^{2}}-\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}=0 ?
$$
Solution:
2865 Upvotes
Verified Answer
The correct answer is:
2
Given differential equation is $\frac{d^{2} y}{d x^{2}}-\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}$
On squaring both the sides,
$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=1+\left(\frac{d y}{d x}\right)^{3}$
Since, degree of the differential equation is the power of highest order derivative. Therefore from above it is clear that degree of equation is 2 .
$\Rightarrow \frac{d^{2} y}{d x^{2}}=\sqrt{1+\left(\frac{d y}{d x}\right)^{3}}$
On squaring both the sides,
$\left(\frac{d^{2} y}{d x^{2}}\right)^{2}=1+\left(\frac{d y}{d x}\right)^{3}$
Since, degree of the differential equation is the power of highest order derivative. Therefore from above it is clear that degree of equation is 2 .
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