# Calculus

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1) yy'-e^x= 0, and y=4 when x=0 means that:
A)y=x-lnx^2+4
B)y^2=4x^2+3
C)y=-2x+1/2x^3
D)y^2=2e^x+14
E)1/2ln|x^2+4|+6

2) The rate of change of P with respect to t is directly proportional to (10-t) and inversely proportional to the cube root of P. The differential equation that models this situation is:
A)dP/dt=k(10-t)/1/3sqrt(p)
B)dP/dt=(10-t)/sqrt(p)
C)dP/dt=k(10-t)3sqrt(p)
D)dP/dt=k3sqrt(p)/10-t
E)dP/dt=k(10-t)/3sqrt(p)

• Calculus -

y dy = e^x dx
1/2 y^2 = e^x + c
if (0,4) is on the graph, then
8 = 1+c
c = 7
1/2 y^2 = e^x + 7
Looks like D to me

dP/dt = k(10-t)/∛t
Looks like E to me
If you must use 3 to indicate the root, why say sqrt? The sq clearly implies square root. Try 3root of cbrt.

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