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Algebra
Question

Again, we are solving for

Again, we are solving for " \( + C ^ { \prime \prime } \) using the given values below, and replacing it into the new integrated function. \( \frac { d y } { d x } = 4 x + 2 , y ( 0 ) = 1 \) 

\( y = 2 x ^ { 2 } + 2 x + 3 \) 

\( y = 2 x ^ { 2 } + 2 x - 1 \) 

\( y = 2 x ^ { 2 } + 2 x + 1 \) 

\( y x ^ { 2 } + 2 x - 2\) 

Answer

y=2x\(^{2}\)+2x+c

1=2(0)+2(0)+c

c=1

y=2x\(^{2}\)+2x+1

Solution
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