-b.2.2.18 IVP DE complete the square?

In summary: The 5 comes from plugging in x=0 into the left side and setting it equal to the right.In summary, the given function can be rewritten as $\frac{dy}{dx}=\frac{e^{-x}-e^x}{3+4y}$ and separated into $3+4y \, dy = e^{-x}-e^x \, dx$. Integrating both sides results in $2y^2+3y=-e^{-x}-e^x+c$. Using the given initial value, the constant can be solved for and the equation can be completed by completing the square. The final solution is $y=\frac{-3+\sqrt{-8e^{-x}-8e^x+65
  • #1
karush
Gold Member
MHB
3,269
5
$\quad\displaystyle
y^{\prime}=
\frac{e^{-x}-e^x}{3+4y},
\quad y(0)=1$
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
$(a)\quad y=-\frac{3}{4}+\frac{1}{4}
+\sqrt{65-8e^x-8e^{-x}}$\\
$(c)\quad|x|<2.0794\textit{ approximately}$
 
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  • #2
karush said:
$\quad\displaystyle
y^{\prime}=
\frac{e^{-x}-e^x}{3+4y},
\quad y(0)=1$
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" ...

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

View attachment 8681

thanks for all the steps

i get lost on the initial value thing

I don't see where the 5 comes from?
 

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  • #4
karush said:
book answer
thanks for all the steps

i get lost on the initial value thing

I don't 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?
 

FAQ: -b.2.2.18 IVP DE complete the square?

What is a complete the square method in IVP DE?

In IVP DE, complete the square is a method for solving differential equations by transforming them into a quadratic equation and using the quadratic formula to find the solution.

Why is it important to complete the square in IVP DE?

Completing the square in IVP DE helps to simplify the differential equation and make it easier to solve. It also helps to identify the initial conditions that are necessary for finding the particular solution.

What are the steps for completing the square in IVP DE?

The steps for completing the square in IVP DE are:
1. Identify the terms with the highest and lowest derivatives.
2. Move the constant term to the right side of the equation.
3. Group the terms with the highest and lowest derivatives together.
4. Add and subtract a constant to the grouped terms to make them perfect squares.
5. Use the quadratic formula to solve for the particular solution.

Can the complete the square method be used for all types of differential equations?

No, the complete the square method is most commonly used for first-order linear differential equations. It can also be used for some second-order linear differential equations, but may not be applicable for more complex equations.

Are there any common mistakes to avoid when using the complete the square method in IVP DE?

Some common mistakes to avoid when using the complete the square method in IVP DE include:
- Forgetting to move the constant term to the right side of the equation
- Incorrectly grouping the terms with the highest and lowest derivatives
- Using the quadratic formula incorrectly
- Not identifying or properly using the initial conditions
- Not simplifying the solution after using the quadratic formula

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