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### WB JEE 2008

MCQ (Single Correct Answer)

The differential equation of all parabolas whose axes are parallel to y-axis is

A
$${{{d^3}y} \over {d{x^3}}} = 0$$
B
$${{{d^2}y} \over {d{x^2}}} = 0$$
C
$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} = 0$$
D
$${{{d^2}y} \over {d{x^2}}} + {{dy} \over {dx}} + y = 0$$

## Explanation

Let equation of parabola whose axis parallel to y-axis, is $${(x - \alpha )^2} = 4a(y - \beta )$$ where ($$\alpha$$, $$\beta$$) is vertex of parabola

differentiating w.r.t. x

$$2(x - \alpha ) = 4a{{dy} \over {dx}}$$ or $$(x - \alpha ) = 2a{{dy} \over {dx}}$$

Taking derivative again

$$1 = 2a{{{d^2}y} \over {d{x^2}}}$$ or $${{{d^2}y} \over {d{x^2}}} = {1 \over {2a}}$$

Taking derivative again

$${{{d^3}y} \over {d{x^3}}} = 0$$.

2

### WB JEE 2008

MCQ (Single Correct Answer)

The degree of the differential equation $${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^{5/3}} = {{{d^2}y} \over {d{x^2}}}$$ is

A
1
B
5
C
10/3
D
3

## Explanation

Given differential equation is

$${\left[ {\left( {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right)} \right]^{5/3}} = {{{d^2}y} \over {d{x^2}}}$$

Cubing both sides

$${\left[ {1 + {{\left( {{{dy} \over {dx}}} \right)}^2}} \right]^5} = {\left( {{{{d^2}y} \over {d{x^2}}}} \right)^3}$$

degree = 3.

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