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[SOLVED] 2..2.18 initial value differentual eq with complete the square??

karush

Well-known member
Jan 31, 2012
2,886
$\quad\displaystyle
y^{\prime}=
\frac{e^{-x}-e^x}{3+4y},
\quad y(0)=1$
rewrite
$\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$
separate
$3+4y \, dy = e^{-x}-e^x \, dx$
integrate
$2y^2+3y=-e^{-x}-e^x+c$
well if so far ok presume complete the square ???


book answer
$(a)\quad y=-\frac{3}{4}+\frac{1}{4}
+\sqrt{65-8e^x-8e^{-x}}$\\
$(c)\quad|x|<2.0794\textit{ approximately}$
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
854
$\quad\displaystyle
y^{\prime}=
\frac{e^{-x}-e^x}{3+4y},
\quad y(0)=1$
rewrite
$\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$
separate
$3+4y \, dy = e^{-x}-e^x \, dx$
integrate
$2y^2+3y=-e^{-x}-e^x+c$
well if so far ok presume complete the square ???


book answer
$(a)\quad y=-\frac{3}{4}+\frac{1}{4}
+\sqrt{65-8e^x-8e^{-x}}$\\
$(c)\quad|x|<2.0794\textit{ approximately}$
$y(0) = 1 \implies 5 = -2+c \implies c = 7$

$y^2 + \dfrac{3}{2}y = \dfrac{-e^{-x} -e^x + 7}{2}$

$y^2 + \dfrac{3}{2}y + \dfrac{9}{16} = \dfrac{-e^{-x} -e^x + 7}{2} + \dfrac{9}{16} $

$\left(y+\dfrac{3}{4}\right)^2 = \dfrac{-8e^{-x}-8e^x+65}{16}$

$y+\dfrac{3}{4} = \dfrac{\sqrt{-8e^{-x}-8e^x+65}}{4}$

check the book "answer" ...

$y = \dfrac{-3 + \sqrt{-8e^{-x}-8e^x+65}}{4}$
 

karush

Well-known member
Jan 31, 2012
2,886
book answer

18.png

thanks for all the steps

i get lost on the initial value thing

I dont see where the 5 comes from?
 
Last edited:

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
854
book answer



thanks for all the steps

i get lost on the initial value thing

I dont see where the 5 comes from?
$2y^2+3y=-e^{-x}-e^x+C$

note $y(0)=1 \implies y = 1 \, \text{when} \, x=0$

$2(1)^2 + 3(1) = -e^{-0}-e^0+C$

... see it now?