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What is a solution to the differential equation #yy'=e^x=0# with y=4 when x=0?

Answer 1

Elijah Abram

# yy'=e^x != 0 #
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Answer 2

Scarlett Allison

To solve the differential equation ( yy' = e^x ), you can separate variables and integrate both sides. The solution is ( y(x) = Ce^x + \frac{1}{2}e^{2x} ), where ( C ) is the constant of integration.

Given the initial condition ( y(0) = 4 ), substitute ( x = 0 ) and ( y = 4 ) into the solution to find ( C ).

( 4 = Ce^0 + \frac{1}{2}e^0 )

( 4 = C + \frac{1}{2} )

( C = \frac{7}{2} )

Therefore, the solution to the differential equation with the given initial condition is ( y(x) = \frac{7}{2}e^x + \frac{1}{2}e^{2x} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.