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What is a solution to the differential equation #ydy/dx=e^x# with y(0)=4?

Answer 1

Christopher Agresta

#implies y^2/2 = e^x + 7#

or if you like...
#implies y = pm sqrt{2(e^x + 7)}#

this is separable and has already been separated

so we integrate both side wrt x

#int ydy/dx \ dx=int e^x \ dx#

#= int y \ dy=int e^x \ dx#

#implies y^2/2 = e^x + C#

applying the IV

#4^2/2 = e^0 + C implies C = 7#

#implies y^2/2 = e^x + 7#

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Answer 2

Axel Alvarez

To solve the differential equation ( \frac{dy}{dx} = e^x ) with the initial condition ( y(0) = 4 ), we can integrate both sides with respect to x.

Integrating ( e^x ) with respect to x gives ( e^x + C ), where C is the constant of integration.

Therefore, the solution to the differential equation is ( y = e^x + C ).

To find the value of C, we use the initial condition ( y(0) = 4 ):

( 4 = e^0 + C ) ( 4 = 1 + C ) ( C = 4 - 1 ) ( C = 3 )

So, the particular solution to the given initial value problem is ( y = e^x + 3 ).

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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.