Question

solve the given initial value problem and determine how the interval in which the solution exists depends on the initial value $y_0$. $$y^{\prime }=2t{y}^2,y(0)=y_0$$

Short answer

Question: Consider the following Initial Value Problem (IVP) involving a first-order differential equation: $y^{\prime }=2t{y}^2$, and the initial value is provided as $y(0)=y_0$. Determine the solution to the IVP and the interval on which the solution exists based on the initial value $y_0$. Answer: The solution to the IVP is given by $y(t)=\frac{1}{\frac{1}{y_0}-t^2}$. The interval of existence for the solution depends on the initial value $y_0$. For $y_0>0$, the interval of existence is $(-\sqrt{\frac{1}{y_0}},\sqrt{\frac{1}{y_0}})$, and for $y_0<0$, the interval of existence is $(-\mathrm{\infty },-\sqrt{\frac{1}{|y_0|}})\cup (\sqrt{\frac{1}{|y_0|}},\mathrm{\infty })$. If $y_0=0$, no interval of existence can be determined as the denominator will always be zero.

Step-by-step solution

Separate Variables

Begin by separating the variables, which will allow us to integrate both sides of the equation. We can rewrite the equation as: $$\frac{dy}{y^2}=2tdt$$

Integrate Both Sides

Now, we will integrate both sides of the equation with respect to their respective variables: $$\int \frac{dy}{y^2}=\int 2tdt$$ Upon integrating, we'll obtain: $$-\frac{1}{y}=t^2+C$$ To find the constant of integration, C, we'll use the given initial condition, $y(0)=y_0$.

Apply Initial Condition and Solve for C

Plug in the initial condition, $y(0)=y_0$, into the equation: $$-\frac{1}{y_0}=(0{)}^2+C$$ From this, we can find the value of C: $$C=-\frac{1}{y_0}$$

Replace C in the Solution and Rewrite the Solution

Now, we'll replace the 'C' in our solution with its value, and rewrite the solution as a function of 'y': $$-\frac{1}{y}=t^2-\frac{1}{y_0}$$ Rearrange to find the function y(t): $$y(t)=\frac{1}{\frac{1}{y_0}-t^2}$$

Determine the Interval of Existence

The solution for 'y(t)', $y(t)=\frac{1}{\frac{1}{y_0}-t^2}$, is valid as long as the denominator is non-zero. Therefore, we have: $$\frac{1}{y_0}-t^2\ne 0$$ Which yields: $$t^2\ne \frac{1}{y_0}$$ In order for the solution to exist, the interval in which 't' can exist depends on the initial value $y_0$: $$t^2<\frac{1}{y_0},y_0>0$$ $$t^2>\frac{1}{y_0},y_0<0$$ For $y_0>0$, the interval of existence is $(-\sqrt{\frac{1}{y_0}},\sqrt{\frac{1}{y_0}})$, and for $y_0<0$, the interval of existence is $(-\mathrm{\infty },-\sqrt{\frac{1}{|y_0|}})\cup (\sqrt{\frac{1}{|y_0|}},\mathrm{\infty })$. Note that if $y_0=0$, no interval of existence can be determined because the denominator will always be zero.

Key concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They play a crucial role in engineering, physics, economics, and other sciences to model the behavior of complex systems. The given exercise involves an initial value problem for a differential equation, which is a special case where you know the value of the function or its derivatives at a certain point.

The equation in our problem, $y^{\text{textprime}}=2t{y}^2$, is an ordinary differential equation (ODE) because it contains one independent variable, $t$, and it is first-order because it involves the first derivative $y^{\text{textprime}}$. Our goal is to find a solution $y(t)$ that satisfies both the equation and the initial condition given by $y(0)=y_0$.

Separation of Variables

Separation of variables is a method commonly used to solve straightforward first-order ODEs, such as the one presented in our exercise. This method involves rearranging the equation so that all terms involving the dependent variable $y$ are on one side, and all terms involving the independent variable $t$ are on the other.

In this case, we separated the variables by dividing by $y^2$ and multiplying by $dt$, resulting in $dy/y^2=2tdt$. This form allows us to integrate each side independently. This step is crucial for turning the differential equation into a solvable algebraic equation.

Integration of Differential Equations

After separating the variables in our ODE, the next step is to integrate both sides. Integration is the inverse operation of differentiation and helps us to find the original function given the derivative.

For the integration step in our problem, we calculate the integral of each side separately, giving $-1/y=t^2+C$, where $C$ is the constant of integration. This constant is determined using the initial condition, which is a distinctive value of the function at a particular point. Here, the initial condition $y(0)=y_0$ helps us find the specific solution to the ODE that we're looking for.

Existence Interval

The existence interval of the solution to an initial value problem indicates where the solution is valid. For a solution to exist, certain conditions must be met. Specifically, the solution and its derivatives must be continuous and finite within the interval.

For the solution $y(t)=1/(\frac{1}{y_0}-t^2)$, the denominator must not be zero. The conditions $t^2<1/y_0$ for $y_0>0$ and $t^2>1/y_0$ for $y_0<0$ define the intervals where the solution exists. The existence interval varies depending on the initial value $y_0$, and it is essential to consider this when analyzing the solution's behavior at different values of $t$.