Question

Find the general solution of the following equations: (a) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=k x\) (b) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=-k y\) (c) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=y^{2}\) (d) \(y \frac{d y}{d x}=\sin x\) (e) \(y \frac{\mathrm{d} y}{\mathrm{~d} x}=x+2\) (f) \(x^{2} \frac{\mathrm{d} y}{\mathrm{~d} x}=2 y^{2}+y x\) (g) \(\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{t^{4}}{x^{5}}\)

Short answer

Question: Solve the following second-order nonhomogeneous linear ordinary differential equations: (a) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} + 2\frac{\mathrm{d} y}{\mathrm{~d} x} + 3y = 2\cos2x\), (b) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} + \frac{\mathrm{d} y}{\mathrm{~d} t} + y = 2\cos3t\), (c) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} + 9y = e^{8x}\), and (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 16x = -9e^{6t}\). Write out the general solution for each of these equations. Answer: (a) \(y(x) = e^{-x}(C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x}) + A\cos{2x} + B\sin{2x}\), where A and B are constants based on the input function \(2\cos2x\). (b) \(y(t) = e^{\frac{-t}{2}}(C_1 \cos{\frac{\sqrt{3}}{2}t} + C_2 \sin{\frac{\sqrt{3}}{2}t}) + A\cos{3t} + B\sin{3t}\), where A and B are constants based on the input function \(2\cos3t\). (c) \(y(x) = C_1 \cos{3x} + C_2 \sin{3x} + Ae^{8x}\), where A is a constant based on the input function \(e^{8x}\). (d) \(x(t) = C_1 e^{4t} + C_2 e^{-4t} + Ae^{6t}\), where A is a constant based on the input function \(-9e^{6t}\).

Step-by-step solution

Find the complementary function

For the homogeneous part of the equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} + 2\frac{\mathrm{d} y}{\mathrm{~d} x} + 3y = 0\), we find the characteristic equation \(r^{2} + 2r + 3 = 0\). Solving this, we get complex roots \(r_1 = -1 + \sqrt{2}i\) and \(r_2 = -1 - \sqrt{2}i\). The complementary function is given by \(y_c = e^{-x}(C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x})\).

Obtain a particular solution

We assume a particular solution in the form \(y_p = A\cos{2x} + B\sin{2x}\). Differentiate twice and substitute into the nonhomogeneous equation to find the coefficients A and B.

General solution

The general solution of the given equation is the sum of the complementary function and the particular solution: \(y(x) = e^{-x}(C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x}) + A\cos{2x} + B\sin{2x}\). ##Equation (b)##

Find the complementary function

For the homogeneous part of the equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} + \frac{\mathrm{d} y}{\mathrm{~d} t} + y = 0\), we find the characteristic equation \(r^2 + r + 1 = 0\). Solving this, we get complex roots \(r_1 = \frac{-1 + i\sqrt{3}}{2}\) and \(r_2 = \frac{-1 - i\sqrt{3}}{2}\). The complementary function is given by \(y_c = e^{\frac{-t}{2}}(C_1 \cos{\frac{\sqrt{3}}{2}t} + C_2 \sin{\frac{\sqrt{3}}{2}t})\).

Obtain a particular solution

We assume a particular solution in the form \(y_p = A\cos{3t} + B\sin{3t}\). Differentiate twice and substitute into the nonhomogeneous equation to find the coefficients A and B.

General solution

The general solution of the given equation is the sum of the complementary function and the particular solution: \(y(t) = e^{\frac{-t}{2}}(C_1 \cos{\frac{\sqrt{3}}{2}t} + C_2 \sin{\frac{\sqrt{3}}{2}t}) + A\cos{3t} + B\sin{3t}\). ##Equation (c)##

Find the complementary function

For the homogeneous part of the equation \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} + 9y = 0\), we find the characteristic equation \(r^2 + 9 = 0\). Solving this, we get complex roots \(r_1 = 3i\) and \(r_2 = -3i\). The complementary function is given by \(y_c = C_1 \cos{3x} + C_2 \sin{3x}\).

Obtain a particular solution

We assume a particular solution in the form \(y_p = Ae^{8x}\). Differentiate twice and substitute into the nonhomogeneous equation to find the coefficient A.

General solution

The general solution of the given equation is the sum of the complementary function and the particular solution: \(y(x) = C_1 \cos{3x} + C_2 \sin{3x} + Ae^{8x}\). ##Equation (d)##

Find the complementary function

For the homogeneous part of the equation \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 16x = 0\), we find the characteristic equation \(r^2 - 16 = 0\). Solving this, we get real roots \(r_1 = 4\) and \(r_2 = -4\). The complementary function is given by \(x_c = C_1 e^{4t} + C_2 e^{-4t}\).

Obtain a particular solution

We assume a particular solution in the form \(x_p = Ae^{6t}\). Differentiate twice and substitute into the nonhomogeneous equation to find the coefficient A.

General solution

The general solution of the given equation is the sum of the complementary function and the particular solution: \(x(t) = C_1 e^{4t} + C_2 e^{-4t} + Ae^{6t}\).

Key concepts

General Solution of Differential Equations

Understanding the general solution of differential equations is crucial for various scientific fields. To put it simply, a differential equation represents a relationship between a function and its derivatives. The 'general solution' of such an equation encompasses all possible solutions that can satisfy the equation.
For instance, in equation (a) \(\frac{\mathrm{d} y}{\mathrm{~d} x}=k x\), the general solution would be of the form \(y = \frac{kx^2}{2} + C\), where \(C\) is an arbitrary constant. Each different value of \(C\) gives us a different particular solution, and together, they form the 'general solution' to the differential equation.
To obtain the general solution, you first find all possible forms of solutions (such as exponential, trigonometric, polynomial, etc.), then combine them, including the arbitrary constants that account for the particular initial or boundary conditions.

Characteristic Equation

The 'characteristic equation' is a key concept when solving linear differential equations with constant coefficients. It translates the differential equation into an algebraic equation, whose solutions determine the form of the complementary function.
For example, in step 1 for equation (b) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}} + \frac{\mathrm{d} y}{\mathrm{~d} t} + y = 0\), the characteristic equation is derived as \(r^2 + r + 1 = 0\). The roots of this equation, complex or real, are essential to forming the complementary function. If the roots are real and distinct, the complementary function has exponential components. If they are complex, as is the case with equation (b), trigonometric components come into play due to Euler's formula.

Complementary Function

The complementary function, often denoted as \(y_c\) or \(x_c\), represents the solution to the associated homogeneous differential equation (the equation with zero on the right-hand side). This function is constructed from the roots of the characteristic equation.
In equation (c) \(\frac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}} + 9y = 0\), the roots are \(r_1 = 3i\) and \(r_2 = -3i\), resulting in the complementary function \(y_c = C_1 \cos{3x} + C_2 \sin{3x}\). Here, \(C_1\) and \(C_2\) are arbitrary constants that would be determined by initial conditions, and their existence is why it's called the 'complementary' function; it complements the particular solution to form the general solution.

Particular Solution

While the complementary function relates to the homogeneous portion of a differential equation, the 'particular solution' is a specific solution that satisfies the non-homogeneous equation.
In problems where the differential equation is non-homogeneous, like in equation (d) \(\frac{\mathrm{d}^{2} x}{\mathrm{~d} t^{2}} - 16x eq 0\), after finding \(x_c\), a particular solution \(x_p\) is assumed. The form of \(x_p\) is based on the nature of the non-homogeneous term. If, for instance, the non-homogeneous term were a polynomial, \(x_p\) would also be assumed as a polynomial. The coefficients in \(x_p\) are then determined by substituting \(x_p\) into the original equation and solving for these unknowns. This solution does not include an arbitrary constant; it is specific to the non-homogeneous part of the equation.

Complex Roots

Complex roots arise in the characteristic equation of a differential equation when there are no real numbers that can satisfy the equation. For example, in the characteristic equation \(r^2 + 2r + 3 = 0\), the roots \(r_1 = -1 + \sqrt{2}i\) and \(r_2 = -1 - \sqrt{2}i\) are complex.
When a differential equation has complex roots, the complementary function contains both exponential and trigonometric terms due to Euler's formula \(e^{ix} = \cos{x} + i\sin{x}\). This results in complementary functions like \(y_c = e^{-x}(C_1 \cos{\sqrt{2}x} + C_2 \sin{\sqrt{2}x})\). Complex roots often lead to oscillatory behavior and are particularly important in the analysis of physical systems like electrical circuits or mechanical vibrations.