A solution of the differential equation {left | vedclass

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Question

(C) The given differential equation is ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$. This can be rewritten as $y = x\frac{{dy

Vedclass · Accepted Answer

$y = 2x - 4$

Vedclass · Answer

(C) The given differential equation is ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$. This can be rewritten as $y = x\frac{{dy}}{{dx}} - {\left( {\frac{{dy}}{{dx}}} \right)^2}$. Let $\frac{{dy}}{{dx}} = p$. Then the equation becomes $y = px - {p^2}$. Differentiating both sides with respect to $x$,we get: $\frac{{dy}}{{dx}} = p + x\frac{{dp}}{{dx}} - 2p\frac{{dp}}{{dx}}$. Since $\frac{{dy}}{{dx}} = p$,we have $p = p + (x - 2p)\frac{{dp}}{{dx}}$. This simplifies to $(x - 2p)\frac{{dp}}{{dx}} = 0$. Case $1$: $\frac{{dp}}{{dx}} = 0$,which implies $p = c$ (a constant). Substituting $p = c$ into $y = px - {p^2}$,we get the general solution $y = cx - {c^2}$. For $c = 2$,we get $y = 2x - {2^2}$,which is $y = 2x - 4$. Thus,$y = 2x - 4$ is a solution.

$A$ solution of the differential equation ${\left( {\frac{{dy}}{{dx}}} \right)^2} - x\frac{{dy}}{{dx}} + y = 0$ is

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