# Solving an Initial Value Problem Analytically

• MHB
• karush
In summary, by solving the initial value problem $y'=\dfrac{1+3x^2}{3y^2-6y}$ with $y(0)=1$ using separation of variables, we obtain the solution $y^3-3y^2=x^3+x+2$ with the restricted interval $|x|<1$ due to the initial condition being at $x=0$. The interval may change depending on the given initial condition.
karush
Gold Member
MHB
Solve the initial value problem
$y'=\dfrac{1+3x^2}{3y^2-6y}, \quad y(0)=1$

Solving analytically
$3y^2-6y\ dy = 1+3x^2 \ dx$

so far hopefully...

Well, I would write it as $(3y^2- 6y)dy= (1+ 3x^2)dx$, but, yes, the differential equation can be "separated" so that y is on one side and x on the other.

Now, integrate both sides. Have you done that?

$y^3 - 3 y^2 =x^3 + x + c$

ok i think we can plug in $x=0$ and $y=1$ so
$1-3=0+0+c$
$c=-2$

the book ans was
$y^3−3y^2−x−x^3+2=0,\quad |x|<1$
but i don't kmow how they got the interval

Last edited:
That requires some careful thought! The original problem was $y'= \frac{1+ 3x^2}{3y^2- 6y}$. The denominator is $3y^2- 6y= 3y(y- 2)$ so y cannot be 0 or 2. You have, correctly, that $y^3- 3y^2- x^3- x+ 2= 0$. If y= 0, that is $-x^3- x+ 2$ which is 0 when x= 1. If y= 2, that is $8- 12- x^4- x+ 2= -x^3- x- 2$ which is 0 when x= -1. Since y cannot be 0 or 2, x cannot be -1 or 1.

Last edited:
mahalo

when I graphed it i noticed vertical slope at 1 but,,,

Notice that the specific condition, |x|< 1, depends also on the fact that the initial condition is at x=0 which is between -1 and 1.

If we had exactly the same differential equation with condition that y(2) be a given value, since 2> 1, we would have to restrict the solution to "x> 1".

If we had exactly the same differential equation with condition that y(-3) be a given value, since -3< -1, we would have to restrict the solution is "x< -1".

## 1. What is an initial value problem?

An initial value problem is a type of mathematical problem where the goal is to find a function that satisfies a given differential equation and a set of initial conditions. The initial conditions typically consist of the value of the function at a specific point or points.

## 2. How is an initial value problem solved analytically?

An initial value problem can be solved analytically by using mathematical techniques such as separation of variables, integrating factors, or the method of undetermined coefficients. These methods involve manipulating the given differential equation to find the general solution, and then using the initial conditions to determine the specific solution that satisfies the problem.

## 3. What are the advantages of solving an initial value problem analytically?

Solving an initial value problem analytically allows for a deeper understanding of the problem and the behavior of the function. It also provides a precise and exact solution, which can be useful in practical applications where accuracy is important.

## 4. Are there any limitations to solving an initial value problem analytically?

Yes, there can be limitations to solving an initial value problem analytically. Some problems may be too complex to be solved analytically, and in those cases, numerical methods may be used instead. Additionally, the analytical solution may not always be feasible or practical to use in real-world situations.

## 5. How can solving an initial value problem analytically be applied in scientific research?

Solving initial value problems analytically is a common approach in scientific research, particularly in fields such as physics, engineering, and economics. It allows researchers to model and analyze complex systems and make predictions about their behavior. Analytical solutions can also serve as a benchmark for comparing and validating numerical methods used in simulations.

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