Problem 31 - Calculus Early Transcendentals

Question

Find the general solution of the following equations. $$\frac{d y}{d t}=\frac{3 t^{2}}{y}$$

Step-by-Step Solution

Answer

Question: Find the general solution of the following first-order differential equation: \(\frac{dy}{dt} = \frac{3t^2}{y}\). Answer: The general solution of the given differential equation is \(y(t) = \pm\sqrt{2t^{3} + 2C}\).

Step 1: Separate the variables

We need to rewrite the equation so that all \(y\) terms are on one side and all \(t\) terms are on the other side. $$y\frac{dy}{dt} = 3t^2$$

Step 2: Integrate both sides

Next, we'll integrate both sides of the equation with respect to their respective variables. $$\int y\frac{dy}{dt} dt= \int 3t^{2} dt$$

Step 3: Evaluate the integral

Solving the integrals on both sides, we get: $$\int y d(y) = \int 3t^{2} dt$$ $$\frac{1}{2}y^{2} = t^{3} + C$$

Step 4: Solve for y

Finally, we need to solve for y to find the general solution. $$y^{2} = 2t^{3} + 2C$$ $$y = \pm\sqrt{2t^{3} + 2C}$$ The general solution of the given differential equation is: $$y(t) = \pm\sqrt{2t^{3} + 2C}$$

Key Concepts

Separation of Variables

In solving differential equations, separation of variables is a method where we rearrange the equation to isolate dependent and independent variables on different sides. This method is especially useful when dealing with separable equations, which have the form \( \frac{dy}{dt} = g(t)h(y) \). By separating variables, we can deal with only one variable at a time during integration.

• This usually involves multiplying or dividing both sides by necessary terms to move all \(y\) variables to one side and all \(t\) variables to the other.
• In our original problem, we separated the equation \(y\frac{dy}{dt} = 3t^2\) into two parts: \(y\) on the left and \(3t^2\) on the right.
• The next step is to integrate both sides with respect to their individual variables.

Separation of variables sets the stage for solving the differential equation by integration in the next step.

Integration

Once the variables are separated, integration is the next step. Integration allows us to find the antiderivative of each side of the equation, essentially 'undoing' the effect of the derivative on each part. This step is crucial because it helps to rebuild the function that forms the solution to the differential equation.

• In our example, after separating variables, we integrate \( \int y \, d(y) \) and \( \int 3t^{2} \, dt \).
• The integral of \( y \) with respect to \( y \) is \( \frac{1}{2}y^2 \), and the integral of \( 3t^2 \) with respect to \( t \) is \( t^3 + C \), where \( C \) is the integration constant.
• These integrals reflect how \( y \) changes with respect to \( t \), taking the form of a new relationship \( \frac{1}{2}y^2 = t^3 + C \).

Integration simplifies complex relationships, making it possible to express one quantity as a function of another.

General Solution

The last step after integration is to solve for the dependent variable to find the general solution. This solution expresses \( y \) in terms of \( t \) while incorporating a constant of integration that accounts for the initial condition and specific cases of the solution.

• Our task is to solve \( \frac{1}{2}y^2 = t^3 + C \) for \( y \). Multiplying through by 2 gives \( y^2 = 2t^3 + 2C \).
• Taking the square root of both sides results in the solutions \( y = \pm \sqrt{2t^3 + 2C} \).
• The "\( \pm \)" symbol indicates that both positive and negative roots are valid solutions, depending on the initial conditions provided with a specific problem scenario.

The general solution provides a family of curves defined by \( C \), which are unique solutions to the differential equation based on different initial conditions.