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Question:

The solution of the differential equation xdy/dx + 2y = x² (x≠0) with y(1) = 1, is?

y=4/5x³+1/5x²

y=3/4x²+1/4x²

y=x²/4+3/4x²

y=x³/5+1/5x²

Solution:

xdy/dx + 2y = x²: y(1) = 1
dy/dx + (2/x)y = x (LDE in y)
IF = e∫(2/x)dx = e^(2lnx) = x²
y * (x²) = ∫x * x² dx = x⁴/4 + C
y(1) = 1
1 = 1/4 + C ⇒ C = 1 - 1/4 = 3/4
yx² = x⁴/4 + 3/4
y = x²/4 + 3/4x².