Asked by Ann
Solve the differential equation dy/dx=3x^2y^2 with the condition that y(1)=4.
I know that y= -1/(x^3 + c) but what do I do with the y(1)=4 part?
I know that y= -1/(x^3 + c) but what do I do with the y(1)=4 part?
Answers
Answered by
Steve
you want to find c. So, since y(1) = 4, plug it in
-1/(1^3+c) = 4
-1/4 = 1+c
c = -5/4
Now the particular solution is
y = -1/(x^3 - 5/4)
-1/(1^3+c) = 4
-1/4 = 1+c
c = -5/4
Now the particular solution is
y = -1/(x^3 - 5/4)
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