Test Index

UPSC NDA Math Model Paper 9

Question : 6 of 120

Marks: +1, -0

\left( \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} \right)^{\frac{1}{6}} = \left( \frac{dy}{dx} + x \right)^{\frac{1}{3}} ?

Solution:

Given that,

\Rightarrow \left( \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} \right)^{\frac{1}{6}} = \left( \frac{dy}{dx} + x \right)^{\frac{1}{3}}

As we can see above equation is not in proper form of polynomial so first we make it in polynomial form,

Multiply by 6 to the both side of power of terms,

\Rightarrow \left( \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} \right)^{\frac{1}{6} \times 6} = \left( \frac{dy}{dx} + x \right)^{\frac{1}{3} \times 6}

\Rightarrow \left( \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} \right)^{1} = \left( \frac{dy}{dx} + x \right)^{2}

\Rightarrow \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} = \left( \frac{dy}{dx} \right)^{2} + 2 \frac{dy}{dx} x + x^{2}

[\because (a+b)^{2}=a^{2}+2ab+b^{2}]

\Rightarrow \frac{d^{2} y}{dx^{2}} + \frac{dy}{dx} - \left( \frac{dy}{dx} \right)^{2} - 2x \cdot \frac{dy}{dx}

-x^{2}=0

Now we can say about the degree and order,

Degree = 1

Order = 2

Go to Question:

Prev Question

Next Question