As per the question,
y=x"\\frac{dy}{dx}"+"\\frac{d^2y}{2dx^2}"
Arranging the above equation as,
"\\frac{d^2y}{dx^2}" +2x"\\frac{dy}{dx}" -2y=0 .....(1)(say)
let y=xt,
we take y=xt since It will fulfill the condition
since we have to reduce Second order differential equation into first order.
differentiate both sides with respect to x,
"\\frac{dy}{dx}" =t
Again differentiating above equation with respect to x,
"\\frac{d^2y}{dx^2}" ="\\frac{dt}{dx}"
Putting the above values in equation 1 we get,
"\\frac{dt}{dx}" +2xt-2xt=0
"\\frac{dt}{dx}" =0
so t=constant=k(say)
since t="\\frac{y}{x}"
"so \\frac{y}{x}" =k
y=kx
This is the required solution.
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