# 48.2.2.23 IVP and min valuew

• MHB
• karush
There is no "c" in the previous line!In summary, the given initial value problem can be solved by separating variables and integrating to obtain the equation $y=-\frac{1}{\frac{x^2}{2}+2x-1}$, or $y=\frac{-2}{x^2+4x-2}$ after multiplying both numerator and denominator by 2. The minimum value can be found by setting the derivative of y to 0, resulting in x = -2. However, there was a typo in the given solution where the constant of integration was incorrectly included in the final equation.
karush
Gold Member
MHB
\tiny{b.48.2.2.23}
Solve the initial value problem $y'=2y^2+xy^2,\quad y(0)=1$
find min value

$\begin{array}{ll} \textit{separate variables} &\dfrac{1}{y^{2}}\ dy=(2+x)\ dx\\ \\ \textit{integrate} & -\dfrac{1}{y}=2x+\dfrac{x^2}{2}+C\\ \\ \textit{plug in$(0,1)$} & -\dfrac{1}{1}=0+0+c \therefore c=-1\\ \\ \textit{the equations is} & y=-1/(x/2+2x-1)+c\\ \\ y'=0 & \textit{returns } x=-2 \end{array}$

typos maybe

There is a typo where you have
$y= -1/(x/2+ 2x- 1)+ c$
You clearly intended "$x^2/2$". Also, having determined that the constant of integration was -1, where did that additional "c" come from?

here is the book answer

yeah your right about the $x^2$

karush said:
\tiny{b.48.2.2.23}
Solve the initial value problem $y'=2y^2+xy^2,\quad y(0)=1$
find min value

$\begin{array}{ll} \textit{separate variables} &\dfrac{1}{y^{2}}\ dy=(2+x)\ dx\\ \\ \textit{integrate} & -\dfrac{1}{y}=2x+\dfrac{x^2}{2}+C\\ \\ \textit{plug in$(0,1)$} & -\dfrac{1}{1}=0+0+c \therefore c=-1\\ \\ \textit{the equations is} & y=-1/(x/2+2x-1)+c\\ \\ y'=0 & \textit{returns } x=-2 \end{array}$

typos maybe
Every thing is good until that last equation!
Integrating gives
$-\frac{1}{y}= \frac{x^2}{2}+ 2x+ C$
an y(0)= 1 gives C= -1 so
$-\frac{1}{y}= \frac{x^2}{2}+ 2x- 1$.

Then solving for y (if you are asked to do so- I would consider the previous line a perfectly good solution)
$y= \frac{-1}{\frac{x^2}{2}+ 2x- 1}$
or, if you don't like a fraction in the denominator of afraction, multiply both numerator and denominator by 2:
$y= \frac{-2}{x^2+ 4x- 2}$

But, again, where did that "+c" in your last line come from?

## 1. What is 48.2.2.23 IVP?

48.2.2.23 IVP stands for "initial value problem" and is a type of mathematical problem that involves finding a solution to a differential equation given an initial condition.

## 2. What is the significance of the min value in 48.2.2.23 IVP?

The min value in 48.2.2.23 IVP refers to the minimum value of the dependent variable (such as time, temperature, or population) in the given initial condition. It helps determine the behavior of the solution to the problem.

## 3. How is 48.2.2.23 IVP solved?

48.2.2.23 IVP is typically solved using mathematical methods such as separation of variables, integrating factors, or Laplace transforms. These methods help find a general solution to the problem, which can then be used to find a specific solution that satisfies the initial condition.

## 4. What are some real-life applications of 48.2.2.23 IVP?

48.2.2.23 IVP has various applications in fields such as physics, engineering, and economics. For example, it can be used to model the growth of a population, the decay of a radioactive substance, or the flow of electricity in a circuit.

## 5. Are there any limitations to using 48.2.2.23 IVP?

48.2.2.23 IVP is a powerful tool for solving differential equations, but it does have some limitations. It may not always be possible to find an exact solution, and numerical methods may need to be used instead. Additionally, the solution may not accurately reflect real-world conditions if the initial conditions are not precise.

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