y+xdxdy=x2
dxdy+xy=x
I.F.=e∫x1dx=x
Solution of differential equation,
y⋅x=∫x⋅xdx
xy=3x3+C
Passes through (−2,2), So,
−12=−8+3C
C=3−4
∴3xy=x3−4
i.e 3x⋅f(x)=x3−4
x3−3xf(x)−4=0
Let us consider a curve, y=f(x) passing through the point (−2,2) and the slope of the tangent to the curve at any point (x,f(x)) is given by f(x)+xf′(x)=x2. Then
Held on 27 Aug 2021 · Verified 6 Jul 2026.
x3−3xf(x)−4=0
x2+2xf(x)−12=0
x3+xf(x)+12=0
x2+2xf(x)+4=0
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