Asked by Anonymous
solve the initial value problem dy/dx = y+2 when y(0)=2
Answers
Answered by
oobleck
no difficulties here...
dy/dx = y+2
dy/(y+2) = dx
ln(y+2) = x + C
Using the point (0,2) to find C, we have
0+C = ln(2+2)
C = ln4
ln(y+2) = x + ln4
y+2 = 4e^x
y = 4e^x - 2
or, ordering the steps differently,
y+2 = e^(x+c)
or, using c = lnC,
y+2 = C*e^x
2+2 = C*e^0 = C
C = 4
y+2 = 4e^x
y = 4e^x - 2
you can easily check to see that this satisfies both conditions.
dy/dx = y+2
dy/(y+2) = dx
ln(y+2) = x + C
Using the point (0,2) to find C, we have
0+C = ln(2+2)
C = ln4
ln(y+2) = x + ln4
y+2 = 4e^x
y = 4e^x - 2
or, ordering the steps differently,
y+2 = e^(x+c)
or, using c = lnC,
y+2 = C*e^x
2+2 = C*e^0 = C
C = 4
y+2 = 4e^x
y = 4e^x - 2
you can easily check to see that this satisfies both conditions.