The differential equation for which l x^2+m y | vedclass

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Question

(A) Given equation is $l x^2 + m y^2 - (x + y) = 0 \ldots (i)$ To eliminate the two arbitrary constants $l$ and $m$, we differentiate the equation twi

Vedclass · Accepted Answer

$\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2x & 2yy^{\prime} & 1+y^{\prime} \ 2 & 2(y^{\prime 2}+yy^{\prime \prime}) & y^{\prime \prime}\end{array}ight|=0$

Vedclass · Answer

(A) Given equation is $l x^2 + m y^2 - (x + y) = 0 \ldots (i)$ To eliminate the two arbitrary constants $l$ and $m$, we differentiate the equation twice with respect to $x$. Differentiating $(i)$ with respect to $x$: $2lx + 2myy^{\prime} - (1 + y^{\prime}) = 0 \ldots (ii)$ Differentiating $(ii)$ with respect to $x$: $2l + 2m(y y^{\prime \prime} + (y^{\prime})^2) - y^{\prime \prime} = 0 \ldots (iii)$ We have a system of three linear equations in $l, m, -1$: $l(x^2) + m(y^2) + (-1)(x+y) = 0$ $l(2x) + m(2yy^{\prime}) + (-1)(1+y^{\prime}) = 0$ $l(2) + m(2(y^{\prime 2} + yy^{\prime \prime})) + (-1)(y^{\prime \prime}) = 0$ For a non-trivial solution for $l, m, -1$, the determinant of the coefficients must be zero: $\left|\begin{array}{ccc}x^2 & y^2 & -(x+y) \ 2x & 2yy^{\prime} & -(1+y^{\prime}) \ 2 & 2(y^{\prime 2} + yy^{\prime \prime}) & -y^{\prime \prime}\end{array}ight| = 0$ Multiplying the third column by $-1$, we get: $\left|\begin{array}{ccc}x^2 & y^2 & x+y \ 2x & 2yy^{\prime} & 1+y^{\prime} \ 2 & 2(y^{\prime 2} + yy^{\prime \prime}) & y^{\prime \prime}\end{array}ight| = 0$

The differential equation for which $l x^2+m y^2=x+y$ is the general solution is

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