Question

Solve the following differential equations: (i) \(x d x=\left(\frac{x^{2}}{y}-y^{3}\right) d y\) (ii) \(\frac{y}{x} d x+\left(y^{3}-\ln x\right) d y=0\) (iii) \(\frac{2 x d x}{y^{3}}+\frac{y^{2}-3 x^{2}}{y^{4}} d y=0\) (iv) \(y-y^{\prime} \cos x=y^{2} \cos x(1-\sin x)\)

Short answer

To summarize: (i) The solution for the first differential equation is: $$x = \ln|y| - \frac{1}{3}y^3 + C$$ (ii) The solution for the second differential equation is: $$- \frac{1}{2y^2} - \ln |\ln x| = C$$ (iii) The solution for the third differential equation is: $$\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{y}{\sqrt{3}x}\right) - \frac{1}{2y^2} = C$$ (iv) The fourth differential equation has no closed-form solution, and further attempts might yield an implicit solution form.

Step-by-step solution

Separate variables

Separate variables by dividing both sides of the equation by \(x\) and \(\left(\frac{x^2}{y} - y^3\right)\): $$\frac{dx}{dy} = \frac{\left(\frac{x^2}{y} - y^3\right)}{x}$$

Integrate both sides

Integrate both sides with respect to \(y\): $$\int \frac{dx}{dy} dy = \int \frac{\left(\frac{x^2}{y} - y^3\right)}{x} dy$$ This gives us: $$x = \int \frac{1}{y} dy - \int y^2 dy + C$$

Solve for x(y)

Simplify the equation and solve for \(x(y)\): $$x = \ln|y| - \frac{1}{3}y^3 + C$$ (ii) \(\frac{y}{x} dx + \left(y^3 - \ln x\right) dy = 0\)

Separate variables

Separate variables by dividing both sides of the equation by \(\left(y^3 - \ln x\right)\) and \(y\): $$\frac{x dy + y dx}{y(xy^3 - x\ln x)} = 0$$ Now, simplify the equation: $$\frac{x dy + y dx}{y^2(xy^3 - x\ln x)} = 0$$ and divide both sides of the equation by \(x\) and \((xy^3 - x\ln x)\): $$\frac{dy}{y^3} - \frac{dx}{\ln x} = 0$$

Integrate both sides

Integrate both sides with respect to \(x\): $$\int \left(\frac{dy}{y^3} - \frac{dx}{\ln x}\right) = 0$$ This gives us: $$- \frac{1}{2y^2} - \ln |\ln x| = C$$ (iii) \(\frac{2x dx}{y^3} + \frac{y^2 - 3x^2}{y^4} dy = 0\)

Take out common term

Take out the common term \(y^3\): $$\frac{2x dx + (y^2 - 3x^2) dy}{y^3} = 0$$

Separate variables

Separate variables by dividing both sides of the equation by \((y^2 - 3x^2)\) and \(2x\): $$\frac{2x dx}{2x(y^2 - 3x^2)} + \frac{(y^2 - 3x^2) dy}{y^3(y^2 - 3x^2)} = 0$$ Now, simplify the equation: $$\frac{dx}{y^2 - 3x^2} + \frac{dy}{y^3} = 0$$

Integrate both sides

Integrate both sides with respect to \(x\): $$\int \left(\frac{dx}{y^2 - 3x^2} + \frac{dy}{y^3}\right) = 0$$ This gives us: $$\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{y}{\sqrt{3}x}\right) - \frac{1}{2y^2} = C$$ (iv) \(y - y^\prime \cos x = y^2 \cos x (1 - \sin x)\)

Rewrite the equation

Rewrite the equation to isolate \(y^\prime\): $$y^\prime = \frac{y - y^2\cos x(1 - \sin x)}{\cos x}$$

Substitute

Introduce substitution \(v(y) = y^2\), such that \(\frac{dv}{dy} = 2y\), or \(\frac{dy}{dv} = \frac{1}{2\sqrt{v}}\): $$\frac{dy'}{dv} = \frac{1}{2\sqrt{v}}$$ And rewrite the equation: $$v^\prime = \frac{\sqrt{v}(1 - v\cos x)}{\cos x}$$

Integrate

Integrate both sides with respect to \(x\): $$\int \frac{dv'}{d x} dx = \int \frac{\sqrt{v}(1 - v\cos x)}{\cos x} dx$$ This equation does not have a closed-form solution, and further attempts might yield an implicit solution form.

Key concepts

Separable Differential Equations

In the realm of mathematics, separable differential equations are a particularly manageable subclass of differential equations that can be solved using simple algebra and integral calculus. They are defined by the ability to separate the variables involved into two sides of the equation, each side containing only one of the variables and its differentials.

Take the exercise provided as an example: the equation \(x dx = \left(\frac{x^2}{y} - y^3\right) dy\) intuitively separates into terms involving 'x' on one side and terms involving 'y' on the other. This separation is pivotal because it enables us to use integration on each variable independently to find a solution.

Integrating the separated equation involves finding the antiderivatives of each side, typically resulting in a general solution that includes an integration constant, denoted as \(C\). It’s important to note that the original equation must be continuous and differentiable for the technique to work successfully, and solutions found may either be explicit or implicit functions of the variables involved.

Integral Calculus

Integral calculus is a cornerstone of mathematics, integral to solving an array of problems including those involving differential equations. It involves the process of finding the integral, or antiderivative, of a function, which represents the accumulation of quantities and can determine areas under curves, among other physical interpretations.

In reference to the earlier exercise, taking the integral of both sides of a separated differential equation essentially ‘reverses’ the differentiation process. For instance, the integration step \(\int \frac{dx}{dy} dy = \int \frac{\left(\frac{x^2}{y} - y^3\right)}{x} dy\) unfolds into \(x = \ln|y| - \frac{1}{3}y^3 + C\). This integration step is an indispensable tool, providing a bridge from the differential world back to the function itself.

While basic functions have well-defined integrals, more complex functions may require special techniques or numerical methods for their evaluation. After integration, the constant 'C' represents the family of solutions to the differential equation, encapsulating the initial conditions or particular solutions into a general form.

First Order Differential Equations

First order differential equations are equations that involve the first derivative of an unknown function and the function itself. Solutions to these types of equations show how a process changes over time, with applications ranging from physics to finance. In the academic context, they serve as an introduction to the broader field of study of differential equations.

Each part of the given homework problems illustrates a varying first order differential equation. Solving these requires not only the separation of variables and integration, as shown in the steps provided, but also an understanding of initial conditions to find specific solutions.

First order differential equations can often be solved analytically, but in cases where they cannot, such as with complex equations or non-separable variables, numerical methods such as Euler's method or Runge-Kutta methods may be employed. Understanding first order differential equations equips students with the fundamental concepts and methods necessary to tackle more advanced differential equations they may encounter in their studies.