y' - y = x^2 y' - y/x = x
Use substitution: u = y/x y = ux y' = xu' + u
xu' + u - y/x = x xu' + u - u = x xu' = x u' = 1
u = ∫ 1 dx u = x + C y/x = x + C y = x^2 + Cx
NOTE: We cannot find for sure that y = x^2 + 3x unless we have some initial condition, such as y(1) = 4Mathmom · 4 years ago1Thumbs up 0Thumbs downCommentReport Abuse
xy' - y = x²
Dividing both sides by x.
y' - (y/x) = x
Now this becomes a first-order linear D.E.
Integrating factor = e^(∫-1/x dx) = 1/x
Multiplying both sides by the integrating factor:
y'/x - (y/x²) = 1
d/dx[y/x] = 1
Integrating both sides:
y/x = x + C
y = x² + Cx
Now you must've had some initial value along with this problem, since you said that the answer is y = 3x + x². This means that C = 3.
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