Solving The Initial Value Problem: A Step-by-Step Guide

Let's dive into solving the initial value problem (IVP) given by:

$$\begin{cases} t y' + 3y + (6t^2 - 3t) y^{4/3} = 0 \\ y(1) = -1 \end{cases}$$

We'll walk through the solution step-by-step to make it clear and easy to follow. Solving differential equations can seem daunting, but with a structured approach, it becomes manageable. Ready? Let's get started!

1. Transforming the Equation

First, we need to transform the given differential equation into a more recognizable form. The equation looks like a Bernoulli equation. To confirm this, let's rewrite the equation as follows:

$$y' + \frac{3}{t}y = -\frac{(6t^2 - 3t)}{t} y^{4/3}$$

Simplify it further to:

$$y' + \frac{3}{t}y = -(6t - 3) y^{4/3}$$

Now it's clearly a Bernoulli equation in the form:

$$y' + P(t)y = Q(t) y^n$$

where $P(t) = \frac{3}{t}$, $Q(t) = -(6t - 3)$, and $n = \frac{4}{3}$.

Why is identifying the form important? Well, recognizing the equation type allows us to apply standard solution techniques, making the problem much easier to handle. In this case, Bernoulli equations have a known method for finding solutions. Essentially, identifying the equation type is the first crucial step towards a successful solution.

2. Applying Bernoulli's Method

To solve the Bernoulli equation, we use the substitution:

$$v = y^{1-n}$$

In our case, $n = \frac{4}{3}$, so:

$$v = y^{1 - \frac{4}{3}} = y^{-\frac{1}{3}}$$

Now, we need to find $y$ in terms of $v$:

$$y = v^{-3}$$

Next, differentiate $y$ with respect to $t$:

$$y' = -3v^{-4} v'$$

Substitute $y$ and $y'$ into the original Bernoulli equation:

$$-3v^{-4} v' + \frac{3}{t} v^{-3} = -(6t - 3) (v^{-3})^{4/3}$$

$$-3v^{-4} v' + \frac{3}{t} v^{-3} = -(6t - 3) v^{-4}$$

Multiply through by $-\frac{v^4}{3}$ to simplify:

$$v' - \frac{1}{t} v = (2t - 1)$$

Now we have a linear first-order differential equation in terms of $v$. Isn't that neat?

3. Solving the Linear First-Order Equation

To solve the linear first-order equation, we need to find the integrating factor $\mu(t)$. The general form of a linear first-order equation is:

$$v' + P(t)v = Q(t)$$

In our case, $P(t) = -\frac{1}{t}$. The integrating factor is:

$$\mu(t) = e^{\int P(t) dt} = e^{\int -\frac{1}{t} dt} = e^{-\ln|t|} = e^{\ln|t^{-1}|} = \frac{1}{t}$$

Now, multiply the entire equation by the integrating factor:

$$\frac{1}{t} v' - \frac{1}{t^2} v = \frac{2t - 1}{t}$$

Notice that the left side is the derivative of $\frac{v}{t}$:

$$\frac{d}{dt} \left( \frac{v}{t} \right) = \frac{2t - 1}{t}$$

Integrate both sides with respect to $t$:

$$\int \frac{d}{dt} \left( \frac{v}{t} \right) dt = \int \frac{2t - 1}{t} dt$$

$$\frac{v}{t} = \int \left( 2 - \frac{1}{t} \right) dt$$

$$\frac{v}{t} = 2t - \ln|t| + C$$

Solve for $v$:

$$v(t) = t(2t - \ln|t| + C)$$

This is the general solution for $v(t)$. Almost there, guys!

4. Applying the Initial Condition

Recall that $v = y^{-\frac{1}{3}}$ and $y(1) = -1$. We need to find the value of $C$ using this initial condition.

$$v(1) = (-1)^{-\frac{1}{3}} = -1$$

Substitute $t = 1$ and $v(1) = -1$ into the equation for $v(t)$:

$$-1 = 1(2(1) - \ln|1| + C)$$

$$-1 = 2 - 0 + C$$

$$C = -3$$

So the particular solution for $v(t)$ is:

$$v(t) = t(2t - \ln|t| - 3)$$

5. Finding y(t)

Now, we need to find $y(t)$ using $v(t) = y^{-\frac{1}{3}}$:

$$y(t) = \left( t(2t - \ln|t| - 3) \right)^{-3}$$

$$y(t) = (2t^2 - t\ln(t) - 3t)^{-3}$$

Thus, the solution to the initial value problem is:

$$y(t) = (2t^2 - t\ln(t) - 3t)^{-3}$$

Make sure to check the solution by plugging it back into the original differential equation and verifying that it satisfies both the equation and the initial condition.

6. Refining and Simplifying

The solution we found is:

$$y(t) = (2t^2 - t\ln(t) - 3t)^{-3}$$

Let's verify whether the provided options match this solution. The given options are:

A. $y = (t^3 - t^2)^{-3}$ B. $y = 2t^2 - 3t - t\ln(t)$

Neither of these directly matches our solution. However, let's analyze them. Option B is not in the form of $y(t)$, so it's likely incorrect. Option A looks like a simplified version but doesn't account for all terms in our solution.

Our derived solution involves a more complex expression inside the parenthesis raised to the power of -3. Hence, it is unlikely that a simple expression like $y = (t^3 - t^2)^{-3}$ would satisfy the given initial value problem.

Therefore, the correct solution remains:

$$y(t) = (2t^2 - t\ln(t) - 3t)^{-3}$$

Important Note: Always double-check your algebra and calculus steps, especially when dealing with initial conditions and substitutions. A small mistake can lead to a completely different answer.

Conclusion

We have successfully solved the initial value problem by recognizing it as a Bernoulli equation, applying the appropriate substitution, solving the resulting linear first-order equation, and using the initial condition to find the particular solution. The final solution is:

$$y(t) = (2t^2 - t\ln(t) - 3t)^{-3}$$

Well done, team! Keep practicing, and you'll become a pro at solving these types of problems. Remember, the key is to break down the problem into manageable steps and stay organized. Happy solving!