lyanalewis
New member
- Joined
- Apr 17, 2010
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- 1
Find curve that passes through (0,2) and has at each point (x,y) a tangent line with slope y - 2e^-x
dy/dx=y-2e^-x
dy/dx-y=-2e^-x
e^(-x)dy/dx-ye^(-x)=-2e^(-2x)
ye^(-x)=Int of -2e^(-2x) dx
ye^(-x)=A+e^(-2x)
y=Ae^x+e^(-x)
y=2 when x=0
A=1
therefore
y=e^x+e^(-x)
then checking
dy/dx=e^x-e^(-x)
dy/dx=y-2e^(-x)
i dont understand starting the bold in aswer. please kindly further explain. thank u
dy/dx=y-2e^-x
dy/dx-y=-2e^-x
e^(-x)dy/dx-ye^(-x)=-2e^(-2x)
ye^(-x)=Int of -2e^(-2x) dx
ye^(-x)=A+e^(-2x)
y=Ae^x+e^(-x)
y=2 when x=0
A=1
therefore
y=e^x+e^(-x)
then checking
dy/dx=e^x-e^(-x)
dy/dx=y-2e^(-x)
i dont understand starting the bold in aswer. please kindly further explain. thank u