Question

Find an integrating factor; that is a function of only one variable, and solve the given equation. $$ 3 x^{2} y d x+2 x^{3} d y=0 $$

Short answer

Question: Solve the following first-order ordinary differential equation (ODE): $$ 3 x^2 ydx + 2 x^3 dy = 0 $$ Answer: The general solution of the given ODE is: $$ y = \frac{3}{2}xe^{2x\big/ 3} + C_1 e^{-2x\big/ 3} $$ where \(C_1\) is an arbitrary constant.

Step-by-step solution

Identify the ODE

We are given the following first-order ODE: $$ 3 x^2 ydx + 2 x^3 dy = 0 $$ Our goal is to find an integrating factor and solve this ODE. #Step 2: Rewrite the ODE in the standard form#

Rewrite the ODE in the standard form

We can rewrite the given ODE in the standard form by dividing both sides by \(3x^2y\). This gives us: $$ dx + \frac{2x}{3y} dy = 0 $$ #Step 3: Find the integrating factor#

Find the integrating factor

We will now find an integrating factor \(\mu(x)\), a function of one variable \(x\), that can make our ODE exact. We can find the integrating factor by taking the partial derivative of the coefficient of \(dy\) with respect to \(x\), and then divide it by the coefficient of \(dx\): $$ \frac{\frac{\partial}{\partial x}\left(\frac{2x}{3y}\right)}{1} = \frac{2}{3y} $$ Now, we can find the integrating factor \(\mu(x)\) by integrating the above equation with respect to \(x\): $$ \mu(x) = e^{\int \frac{2}{3y} dx} = e^{2x\big/ 3} $$ #Step 4: Multiply the ODE by the integrating factor#

Multiply the ODE by the integrating factor

Now, we will multiply the given ODE by the integrating factor \(\mu(x)\): $$ e^{2x\big/ 3} (dx + \frac{2x}{3y} dy) = 0 $$ #Step 5: Check if the ODE is exact#

Check if the ODE is exact

The ODE will be exact if the following condition is satisfied: $$ \frac{\partial}{\partial y} \left(e^{2x\big/ 3}\right) = \frac{\partial}{\partial x} \left(\frac{2x}{3y} e^{2x\big/ 3}\right) $$ By calculating the partial derivatives, we find that the above condition is satisfied, so our ODE is now exact. #Step 6: Integrate the ODE and find the general solution#

Integrate the ODE and find the general solution

Since the ODE is exact, we can now integrate both sides to find the general solution: $$ \int e^{2x\big/ 3} dx + \int \frac{2x}{3y} e^{2x\big/ 3} dy = \int 0 dx $$ By integrating, we obtain: $$ \frac{3}{2}e^{2x\big/ 3} + \frac{2}{3} x e^{2x\big/ 3} + C(y) = C $$ Finally, to express \(y\) as a function of \(x\), we solve for \(y\): $$ y = \frac{3}{2}xe^{2x\big/ 3} + C_1 e^{-2x\big/ 3} $$ Where \(C_1\) is an arbitrary constant. This is the general solution of the given ODE.

Key concepts

Integrating Factor

An integrating factor is a mathematical tool often used to solve first-order linear ordinary differential equations (ODEs). For an ODE that is not exact, which means it cannot be solved by simple integration of both sides, an integrating factor can sometimes transform it into an exact equation.

An integrating factor is typically denoted as \( \text{\(\mu(x)\)} \) or \( \text{\(\mu(y)\)} \), depending on whether it is a function of \(x\) or \(y\). The beauty of this method lies in its ability to multiply the entire ODE by \( \text{\(\mu\)} \), making the resultant expression a perfect differential that can be integrated straightforwardly.

Application to the Exercise

In our exercise, we identified an integrating factor for the ODE \(3 x^{2} y dx + 2 x^{3} dy = 0\). By finding \( \text{\(\mu(x)\)} \) and multiplying it across the ODE, we converted the non-exact differential equation into an exact one, paving the way for finding a solution.

Exact Differential Equations

An exact differential equation is a specific type of ODE where there exists a function \(U(x, y)\) such that its total differential matches the given equation. That means, for an ODE of the form \(M(x, y)dx + N(x, y)dy = 0\), it is exact if there exists \(U(x, y)\) with \(\frac{\partial U}{\partial x} = M(x, y)\) and \(\frac{\partial U}{\partial y} = N(x, y)\).

An exact differential equation implies that the expression on the left side of the equation comes directly from the gradient of a scalar field

• If \(M_y = N_x\), then our equation is exact.

Application to the Exercise

In this exercise, after an integrating factor was applied, we verified the exactness by checking the cross partial derivatives and confirmed that the condition for exactness was satisfied, allowing us to proceed to integrate and find a solution.

Partial Derivatives

The concept of partial derivatives is an essential tool in the realm of multivariable calculus. A partial derivative represents the rate at which a function changes as one variable changes while all other variables are held constant. It is denoted by \(\frac{\partial}{\partial x}\) or \(\frac{\partial}{\partial y}\), depending on the variable of differentiation.

Partial derivatives are crucial when dealing with functions of more than one variable and are fundamental in determining the exactness of an ODE, optimization problems, and more. They are the building blocks for more advanced concepts in vector calculus and differential geometry.

Application to the Exercise

In solving our ODE, we used partial derivatives to find the integrating factor and then to verify the exactness. Through partial differentiation of the coefficients of \(dx\) and \(dy\), we ensured that the modified equation was exact, which was essential to find the general solution.