differential equation

Question

Find the particular solution that satisfies the differential equation  with the initial condition 

Answer 1

Separate variables:

ydy=e^xdx

Then:

(1/2)y²=e^x+C₁

y²=2e^x+C₂

When x = 0, y = 4:

16=2+C₂

C₂=14

y=√(2e^x+14) --using positive square root, otherwise x = 0 gives y = -4.

Check: y' = e^x/(√(2e^x+14))

yy'-e^x= 0

Here is a Desmos graph for verification:

desmos.com/calculator/vwskvnztqd

And another as part of a slope field:

desmos.com/calculator/1cv9bym84g

Answer 2

have y(dy/dx)=e^x

y(dy)=e^x(dx)

∫y(dy)=∫e^x(dx)

(y²)/2=e^x+C

x=0,,,y=4,,,C=7

y²=2(e^x+7)

as Doug C. points out correctly

since y=+4 when x=0

y=+√(2(e^x+7))