Q.3 (C) (ii) Obtain the differential equation from the relation Ax²+By²=1 , where A and B are constants. Solution : The given equation is Ax²+By²=1 .....(I) Differentiating equation (I) w.r.t. x, we get square x+2Bydy/dx=0 , i.e. Ax+Bydy/dx=0 .....(II) Differentiating equation (II) w.r.t. x, we get A+B square=0 .....(III) Since equations (I), (II) and (III) are consistent in A and B, beginvmatrixx² & y² & 1\\ x & ydy/dx & 0\\ 1 & square & 0endvmatrix=0 therefore \x[yd²y/dx²+(dy/dx)²]-ydy/dx\=0 therefore square+x(dy/dx)²-ydy/dx=0

Answer: xyd²y/dx²+x(dy/dx)²-ydy/dx=0 Completed boxes: First derivative: 2Ax+2Bydy/dx=0 , so the box is 2A . Differentiating Ax+Bydy/dx=0 : A+B[yd²y/dx²+(dy/dx)²]=0 , so the box in (III) and in the determinant is yd²y/dx²+(dy/dx)² . Expanding the determinant along the third column: 1·\x[yd²y/dx²+(dy/dx)²]-ydy/dx\=0 , so the last box is xyd²y/dx² . The required differential equation is xyd²y/dx²+x(dy/dx)²-ydy/dx=0 .

Obtain the differential equation from the relation \(Ax^2+By^2=1\), where A and B are constants.Solution : The given equation is \(Ax^2+By^2=1\) .....(I)Differentiating equation (I) w.r.t. x, we get \(\square\,x+2By\frac{dy}{dx}=0\), i.e. \(Ax+By\fra…

Q.3 (C) (ii) Obtain the differential equation from the relation \(Ax^2+By^2=1\), where A and B are constants.

Solution : The given equation is \(Ax^2+By^2=1\) .....(I)

Differentiating equation (I) w.r.t. x, we get \(\square\,x+2By\frac{dy}{dx}=0\), i.e. \(Ax+By\frac{dy}{dx}=0\) .....(II)

Differentiating equation (II) w.r.t. x, we get \(A+B\,\square=0\) .....(III)

Since equations (I), (II) and (III) are consistent in A and B, \(\begin{vmatrix}x^2 & y^2 & 1\\ x & y\frac{dy}{dx} & 0\\ 1 & \square & 0\end{vmatrix}=0\)

\(\therefore \left\{x\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]-y\frac{dy}{dx}\right\}=0\)

\(\therefore \square+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0\)

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Answer: \(xy\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0\)

Completed boxes:

First derivative: \(2Ax+2By\frac{dy}{dx}=0\), so the box is \(2A\).
Differentiating \(Ax+By\frac{dy}{dx}=0\): \(A+B\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]=0\), so the box in (III) and in the determinant is \(y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\).
Expanding the determinant along the third column: \(1\cdot\left\{x\left[y\frac{d^2y}{dx^2}+\left(\frac{dy}{dx}\right)^2\right]-y\frac{dy}{dx}\right\}=0\), so the last box is \(xy\frac{d^2y}{dx^2}\).

The required differential equation is \(xy\frac{d^2y}{dx^2}+x\left(\frac{dy}{dx}\right)^2-y\frac{dy}{dx}=0\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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Obtain the differential equation from the relation \(Ax^2+By^2=1\), where A and B are constants.Solution : The given equation is \(Ax^2+By^2=1\) .....(I)Differentiating equation (I) w.r.t. x, we get \(\square\,x+2By\frac{dy}{dx}=0\), i.e. \(Ax+By\fra…

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