solve the differentiatial equation y'=x+y?

Question

How do you solve the differentiatial equation y'=x+y by making the change of variable u=x+y?

Mathmom · Accepted Answer

u = x+y
du/dx = 1 + 1*dy/dx
dy/dx = du/dx - 1

y' = x+y
dy/dx = x+y
du/dx - 1 = u
du/dx = u+1
du = (u+1)dx
du/(u+1) = dx

Integrate both sides:
log(u+1) = x + c
log(x+y+1) = x+c
(x+y+1) = e^(x+c) = e^x * e^c

Since c is an arbitrary constant, we can replace e^c with k, another arbitrary constant.

(x+y+1) = ke^x

ANSWER: y = ke^x - x - 1

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Check:
y = ke^x - x - 1
y' = ke^x - 1

x+y = x + ke^x - x - 1 = ke^x - 1

therefore
y' = ke^x - 1 = x+y

solve the differentiatial equation y'=x+y?

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