integrating factor in finding curve

[lyanalewis]

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

 

[BigGlenntheHeavy]

$\displaystyle y_1 \ = \ e^{x}+e^{-x}, \ correct$

$\displaystyle y_1' \ = \ e^{x}-e^{-x} \ = \ m \ = \ y_2-2e^{-x}$

$\displaystyle Hence, \ y_2-2e^{-x} \ = \ e^{x}-e^{-x}, \ y_2 \ = \ e^{x}+e^{-x} \ = \ y_1$

$\displaystyle Therefore, \ y_2 \ = \ m \ +2e^{-x} \ = \ y_1 \ = \ e^{x}+e^{-x}$

$\displaystyle So, \ m \ +2e^{-x} \ = \ e^{x}+e^{-x}, \ m \ =e^{x}-e^{-x} \ = \ y_1', \ QED$

 

[BigGlenntheHeavy]

$\displaystyle y(x) \ = \ e^x+e^{-x}, \ y(0) \ = \ e^0+e^{-0} \ = \ 1+1 \ = \ 2, \ checks.$

$\displaystyle y'(x) \ = \ e^x-e^{-x}, \ Now, \ say \ let \ x \ = \ 2, \ then \ y \ = \ e^2+e^{-2}$

$\displaystyle thus, \ (2,e^2+e^{-2}) \ is \ a \ point \ on \ the \ curve \ y(x) \ = \ e^x+e^{-x}.$

\(\displaystyle Now, \ all \ we \ need \ is \ the \ slope \ at \ this \ point \ on \ y \ and \ then \ we \ will \ have \\)

$\displaystyle enough \ information \ to \ draw \ the \ line \ tangent \ to \ the \ curve \ at \ this \ point.$

$\displaystyle y'(x) \ = \ e^x-e^{-x}, \ hence \ y'(2) \ = \ e^2-e^{-2} \ = \ m, \ the \ slope \ when \ x \ = \ 2 \ on \ the \ curve.$

$\displaystyle OK. \ now \ we \ are \ in \ business \ as \ y-(e^2+e^{-2}) \ = \ (e^2-e^{-2})(x-2) \ or$

$\displaystyle y \ = \ (e^2-e^{-2})(x-2)+(e^2+e^{-2}), \ see \ graph.$

$\displaystyle Now, \ we \ want \ to \ know \ does \ y-2e^{-x} \ equal \ the \ slope \ when \ x \ = \ 2 \ and \ y \ = \ e^2+e^{-2}?$

$\displaystyle Well, \ let's \ check: \ y-2e^{-x} \ = \ e^2+e^{-2}-2e^{-2} \ = \ e^2-e^{-2}, \ QED$

$\displaystyle Post \ Script: \ Thank \ you \ Herr \ Leonhard \ Euler \ for \ e.$

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