Differential Equations

The order of a differential equation is the highest order of the derivative that appears in the equation. In this case, the highest order of the derivative is 3, so the order of the differential equation is 3.

The degree of a differential equation is the highest power of the highest order derivative that appears in the equation. In this case, the highest power of the highest order derivative is 2, so the degree of the differential equation is 2.

To solve this differential equation, we can use the method of separation of variables. Rewriting the equation as $\frac{dy}{y} = \frac{dx}{x}$, we can integrate both sides to get $\ln y = \ln x + C$, where $C$ is the constant of integration. Exponentiating both sides, we get $y = Ce^x$, which is the general solution. Since $C = e^C$, we can write the general solution as $y = Cx$, where $C$ is an arbitrary constant.

To solve this differential equation, we can use the method of characteristic equations. The characteristic equation is $r^2 - 4r + 4 = 0$, which has two roots $r_1 = r_2 = 2$. Therefore, the general solution is $y = C_1 e^{2x} + C_2 e^{-2x}$, where $C_1$ and $C_2$ are arbitrary constants.

To solve this differential equation, we can use the method of undetermined coefficients. Since the right-hand side is $\sin x$, we guess that the particular solution has the form $y_p = A \sin x + B \cos x$. Substituting this into the differential equation, we get $A \sin x + B \cos x - A \sin x - B \cos x = \sin x$, which implies that $A = \frac{1}{2}$ and $B = 0$. Therefore, the particular solution is $y_p = \frac{1}{2} \sin x$. The general solution is then $y = C_1 \cos x + C_2 \sin x + y_p = C_1 \cos x + C_2 \sin x + \frac{1}{2} \sin x$.

To solve this differential equation, we can use the method of partial fractions. Rewriting the equation as $\frac{dy}{y(y-1)(y-2)} = dx$, we can decompose the fraction into partial fractions as $\frac{1}{y(y-1)(y-2)} = \frac{A}{y} + \frac{B}{y-1} + \frac{C}{y-2}$. Solving for $A$, $B$, and $C$, we get $A = 1$, $B = -1$, and $C = 1$. Therefore, the differential equation becomes $\frac{dy}{y} - \frac{dy}{y-1} + \frac{dy}{y-2} = dx$. Integrating both sides, we get $\ln y - \ln (y-1) + \ln (y-2) = x + C$, where $C$ is the constant of integration. Simplifying this expression, we get $\ln \frac{y(y-2)}{y-1} = x + C$, which implies that $y = \frac{1}{C_1 e^x - 1}$, where $C_1 = e^C$.

To solve this differential equation, we can use the method of Laplace transforms. Taking the Laplace transform of both sides of the differential equation, we get $s^2 Y(s) - s y(0) - y'(0) + 4Y(s) = 1$, where $Y(s)$ is the Laplace transform of $y(t)$. Solving for $Y(s)$, we get $Y(s) = \frac{1}{s^2 + 4} + \frac{s y(0) + y'(0)}{s^2 + 4}$. Taking the inverse Laplace transform of $Y(s)$, we get $y(t) = \frac{1}{2} \sin 2t + C_1 \cos 2t + \frac{1}{2}$, where $C_1$ is an arbitrary constant.

To solve this differential equation, we can use the method of separation of variables. Rewriting the equation as $\frac{dy}{y^2} = dx$, we can integrate both sides to get $\frac{-1}{y} = x + C$, where $C$ is the constant of integration. Solving for $y$, we get $y = \frac{1}{C_1 - x}$, where $C_1 = -C$.

To solve this differential equation, we can use the method of separation of variables. Rewriting the equation as $\frac{dy}{y} = \frac{dx}{x}$, we can integrate both sides to get $\ln y = \ln x + C$, where $C$ is the constant of integration. Exponentiating both sides, we get $y = Ce^x$, which is the general solution. Since $C = e^C$, we can write the general solution as $y = Cx$, where $C$ is an arbitrary constant.

To solve this differential equation, we can use the method of characteristic equations. The characteristic equation is $r^2 - 4r + 4 = 0$, which has two roots $r_1 = r_2 = 2$. Therefore, the general solution is $y = C_1 e^{2x} + C_2 e^{-2x}$, where $C_1$ and $C_2$ are arbitrary constants.

To solve this differential equation, we can use the method of undetermined coefficients. Since the right-hand side is $\sin x$, we guess that the particular solution has the form $y_p = A \sin x + B \cos x$. Substituting this into the differential equation, we get $A \sin x + B \cos x - A \sin x - B \cos x = \sin x$, which implies that $A = \frac{1}{2}$ and $B = 0$. Therefore, the particular solution is $y_p = \frac{1}{2} \sin x$. The general solution is then $y = C_1 \cos x + C_2 \sin x + y_p = C_1 \cos x + C_2 \sin x + \frac{1}{2} \sin x$.

To solve this differential equation, we can use the method of partial fractions. Rewriting the equation as $\frac{dy}{y(y-1)(y-2)} = dx$, we can decompose the fraction into partial fractions as $\frac{1}{y(y-1)(y-2)} = \frac{A}{y} + \frac{B}{y-1} + \frac{C}{y-2}$. Solving for $A$, $B$, and $C$, we get $A = 1$, $B = -1$, and $C = 1$. Therefore, the differential equation becomes $\frac{dy}{y} - \frac{dy}{y-1} + \frac{dy}{y-2} = dx$. Integrating both sides, we get $\ln y - \ln (y-1) + \ln (y-2) = x + C$, where $C$ is the constant of integration. Simplifying this expression, we get $\ln \frac{y(y-2)}{y-1} = x + C$, which implies that $y = \frac{1}{C_1 e^x - 1}$, where $C_1 = e^C$.

The order of a differential equation is determined by the highest order of the derivative that appears in the equation.

The general solution of a differential equation is a solution that contains arbitrary constants, which can be determined by applying initial or boundary conditions.

An integrating factor is a function that, when multiplied by a differential equation, makes the equation exact, meaning that it can be integrated directly.