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solve the following differential equation f''(x)=2 , f'(1)=4 , f(2) =-2
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f''(x) = 2 int(f''(x)) = int(2) ... Integrate both sides f'(x) = 2x + C Since f'(1) = 4, we can say 4 = 2(1) + C 4 = 2 + C 2 = C C = 2 So f'(x) = 2x + 2 f'(x) = 2x+2 int(f'(x)) = int(2x+2) f(x) = x^2 + 2x + C Because f(2) = -2, we know that -2 = (2)^2 + 2(2) + C -2 = 4 + 4 + C -2 = 8 + C -10 = C C = -10 So f(x) = x^2 + 2x - 10
thank you
sure thing
I loved how explained it to me lol
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