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Find the function y(x) satisfying dy/dx=8x-3 and y(2)=0y(x)=?

User D Hansen
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1 Answer

1 vote

Answer:


y(x)=4x^2 -3x -10

Explanation:

First take the integral of dy/dx to find a general formula for y(x):


(dy)/(dx) =8x-3\\y(x) = \int{(8x-3)} \, dx =4x^2 -3x +c

Then, evaluate the expression found above at y(2)=0 in order to find the value for the constant 'c':


y(x) = 4x^2 -3x +c\\y(2) = 0\\0= 4*(2)^2 -(3*2) +c\\c=6-16 = -10

The expression for y(x) that satisfies both conditions is:


y(x)=4x^2 -3x -10

User Jns
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