
Yefim S. answered 10/31/20
Math Tutor with Experience
Dividing by ydx we get; dy/dx = 2x(y + 1)/y; ydy/(y + 1) = 2xdx; ∫ydy/(y + 1) = ∫2xdx; ∫(1 - 1/(y + 1))dy = x2 + C;
y - ln(1 + y) = x2 + C. This is general solution in implicit form
Taha T.
asked 10/31/20Find the general solution of the following equation?
y dy - 2 (yx+x) dx = 0
Please write the solution with full steps
Yefim S. answered 10/31/20
Math Tutor with Experience
Dividing by ydx we get; dy/dx = 2x(y + 1)/y; ydy/(y + 1) = 2xdx; ∫ydy/(y + 1) = ∫2xdx; ∫(1 - 1/(y + 1))dy = x2 + C;
y - ln(1 + y) = x2 + C. This is general solution in implicit form
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Taha T.
thanks a lot, but why did we change y/(y+1) to (1-1/y+1))? :( cuz my teacher did the same thing but i didn't understand why.11/18/20