# -b.2.2.26 IVP min value y'=2(1+x)(1+y^2); y(0)=0

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• karush
In summary, To solve the IVP $y'=2(1+x)(1+y^2),\quad y(0)=0$, we first separate the variables and integrate through to get $\arctan(y)=2x+x^2+c$. Plugging in the given initial conditions, we find that $c=0$ and thus the equation is $y=\tan(2x-x^2)$. To find where the solution attains its minimum value, we set the derivative equal to 0 and solve for x, giving us $x=-1$. To check whether this is a minimum or maximum, we use the second derivative test and find that it is indeed a minimum.
karush
Gold Member
MHB
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Solve the IVP $y'=2(1+x)(1+y^2),\quad y(0)=0$
$\begin{array}{ll} \textit{separate variables}& \displaystyle \left(\dfrac{1}{1+y^2}\right)\ dy =2(1+x)\ dx\\ \textit{integrate thru}& \arctan \left(y\right)=2x+x^2+c\\ \textit{plug in x=0 and y=0}& \arctan 0=0+c\\ &0=c\\ \textit{thus the equation is}& y=\tan(2x-x^2) \end{array}$
find where solution attains minimum value
ok wasn't sure about the tangent thing
book answer for min value is -1 but ?

Last edited:
By the time you are studying differential equations, you should know that the standard method for finding a minimum, or maximum, value for a function is to set the derivative equal to 0.

Here, you are told immediately that the derivative is $y'= 2(1+ x)(1+ y^2)$. Setting that equal to 0, $2(1+ x)(1+ y^2)= 0$. Since $1+ y^2$ is never 0 we must have x= -1.

Add: To check whether x= -1 gives a max or min, use the "second derivative test". $y'= 2(x+ 1)(1+ y^2)$ so $y''= 2(1+ y^2)+ 2(1+ x)(2yy')$. At x= -1 that is $y''= 2(1+ y(-1)^2)$ which is positive no matter what y(-1) is. Therefore x= -1 gives a minimum.

ok ill try that on the next one

## 1. What is the meaning of "IVP" in the given equation?

"IVP" stands for "initial value problem". In this context, it means that we are given an initial condition (y(0)=0) and we need to find the solution to the differential equation that satisfies this condition.

## 2. What does "min value" refer to in this equation?

"Min value" refers to the minimum value of the solution to the differential equation. In other words, we are looking for the smallest possible value that y can take on while still satisfying the given equation and initial condition.

## 3. How do you solve this type of differential equation?

This is a separable differential equation, which means we can separate the variables (y and x) and integrate both sides to find the solution. The general steps for solving separable differential equations are: 1) separate the variables, 2) integrate both sides, and 3) solve for y.

## 4. What is the significance of the constants in this equation?

The constants (2, 1, and 0) in this equation represent the coefficients of the variables (x and y) and the initial condition, respectively. They help us to find the specific solution to the differential equation that satisfies the given initial condition.

## 5. Can this equation be solved analytically or numerically?

Yes, this equation can be solved both analytically (using the general steps for solving separable differential equations) and numerically (using numerical methods such as Euler's method or Runge-Kutta methods). The choice of method depends on the specific problem and the desired level of accuracy.

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