The solution of the differential equation p-x²=q+y² \; { is } \;

Correct Answer is : z=(x33+cx)+(cy−y33)+c1z = ≤ft( {x³}{3} + c x ) + ≤ft( c y - {y³}{3} ) + c₁z=(3x3+cx)+(cy−3y3)+c1

z = ≤ft( {x³}{3} + c x ) + ≤ft( c y + {y³}{3} ) + c₁

z = ≤ft( {x³}{3} - x ) + ≤ft( c y - {y³}{3} ) + c₁

z = ≤ft( {x³}{3} - c x ) + ≤ft( c y + {y³}{3} ) + c₁

z = ≤ft( {x³}{3} + c x ) + ≤ft( c y - {y³}{3} ) + c₁

Correct Answer is : z=(x33+cx)+(cy−y33)+c1z = ≤ft( {x³}{3} + c x ) + ≤ft( c y - {y³}{3} ) + c₁z=(3x3+cx)+(cy−3y3)+c1

Test Index

UPSC ESE 2025 Electronics and Telecommunications Paper

Question : 143 of 150

Marks: +1, -0

p-x^{2}=q+y^{2} \; \text{ is } \;

$z = \left( \frac{x^{3}}{3} + c x \right) + \left( c y + \frac{y^{3}}{3} \right) + c_{1}$
$z = \left( \frac{x^{3}}{3} - x \right) + \left( c y - \frac{y^{3}}{3} \right) + c_{1}$
$z = \left( \frac{x^{3}}{3} - c x \right) + \left( c y + \frac{y^{3}}{3} \right) + c_{1}$
$z = \left( \frac{x^{3}}{3} + c x \right) + \left( c y - \frac{y^{3}}{3} \right) + c_{1}$

Solution:

(d)

Given

D

equation is

p-x^{2}=q+y^{2}

p-q=x^{2}+y^{2}

\begin{equation*} \frac{\partial z}{\partial x} - \frac{\partial z}{\partial y} = x^{2}+y^{2} \tag{1} \end{equation*}

Taking option (d) is solution,

z = \left( \frac{x^{3}}{3} + c x \right) + \left( c y - \frac{y^{3}}{3} \right) + c_{1}

p = \frac{\partial z}{\partial x} = x^{2}+c

q = \frac{\partial z}{\partial y} = c - y^{2}

Now

p - q = (x^{2}+c) - (c - y^{2}) = x^{2}+y^{2}

Which is equation (1)

Hence our assumptions is correct.

Hence, the correct option is (d).

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