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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.