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Order of a Differential Equation

Find the orde...

Question

Find the order and degree of $[1 + {y^{'}}^{2}]^{1 / 2} = x^{2} + y$

1, 2

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2, 1

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1, 1

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2, 2

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Solution

The correct option is A $1, 2$
${[1 + {(y^{'})}^{2}]}^{1 / 2} = x^{2} + y$ in this equation

The highest derivative is the first order
And the power of highest order derivative is 2
Hence order is 1 and the degree is 2

Suggest Corrections

Similar questions

(1 + y^{'})^{1 / 2} = y^{''}

x^{2} + y^{2} + z^{2} = 1

y z + z x + x y

y^{'''} + y^{2} + e^{y^{1}} = 0

A = {1, 2}

B = {2, 3, 4}

A \times A = {(1, 1), (1, 2), (2, 1), (2, 2)}

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A = {x : x \in Z, | x | \leq 3}

B = {y; y \in N, - 1 < y + 2 \leq 4}

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