Next We Have the Equation
y ′ = 6 x 2 + x − 5
Taking as initial condition y ( 0 ) = 2
Find y ( 4 )
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Nice. But I like my solution. :D
First let us take the antiderivative of both sides. This evaluates to y = 2 x 3 + 2 x 2 − 5 x + C . We can plug in zero to this equation and set it equal to two to solve for C . Thus, 0 + C = 2 and C equals two. From there, with our general form, we can plug in four and get 1 1 8 .
Very Easy , I'm Starting With Differential Equations !
In Soon I Hope Make Hard Questions !
Vote Up For You !!
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At least give this problem rating. I hate when answering a problem but get nothing. :D
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My problems Are All Easy!
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@Gabriel Merces – Uh... do you need to capitalize all of your words?
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@Finn Hulse – @Finn Hulse Gabriel is a bit newbie at english,, So he has such minor errors....
@Gabriel Merces – Please don't use punctuation mark in the end of your comment. It sounds you are angry. :)
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By the fundamental theorem of calculus,
f ( 4 ) = f ( 0 ) + ∫ 0 4 f ′ ( x ) d x = 2 + ∫ 0 4 ( 6 x 3 + x − 5 ) d x = 1 1 8