Ordinary Differential Equations

--- Given data: $x^2 y' + xy = 1$, $\; y(1) = 2$, $\; x 0$ Divide by $x^2$: $$\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x^2}$$ So $P = \dfrac{1}{x}$, $Q = \dfrac{1}{x^2}$. $$\text{I.F.} = e^{\int \frac{1}{x}\,dx} = e^{\ln x} = x$$ $$y\cdot x = \int \frac{1}{x...

Given: - Differential equation: $y' + 2xy - 1 = 0$ No initial condition given, so we seek the general solution. --- $$\frac{dy}{dx} + 2xy = 1$$ This is a first-order linear ODE: $$\frac{dy}{dx} + P(x)\,y = Q(x), \qquad P(x) = 2x, \quad Q(x) = 1$$ $$\text{I....

- Family of functions: $y = \dfrac{1 + ce^t}{1 - ce^t}$ - Differential equation: $y' = \dfrac{1}{2}(y^2 - 1)$ Using the quotient rule with $u = 1 + ce^t$, $v = 1 - ce^t$: $$u' = ce^t, \qquad v' = -ce^t$$ $$y' = \frac{u'v - uv'}{v^2} = \frac{ce^t(1 - ce^t) -...

- Proposed solution: $y = x - \dfrac{1}{x}$ - Differential equation: $xy' + y = 2x$ --- $$y = x - \frac{1}{x} = x - x^{-1}$$ $$y' = \frac{d}{dx}(x) - \frac{d}{dx}(x^{-1}) = 1 - (-1)x^{-2} = 1 + \frac{1}{x^2}$$ --- $$xy' = x\left(1 + \frac{1}{x^2}\right) = x...

Part 1: ODE $y' = \dfrac{x^2}{y^2}$, initial condition $y(0)=2$. Part 2: ODE $y'' + y' - 6y = 0$, initial conditions $y(0)=0$, $y'(0)=1$. --- $$\frac{dy}{dx} = \frac{x^2}{y^2} \implies y^2 \, dy = x^2 \, dx$$ $$\int y^2 \, dy = \int x^2 \, dx$$ $$\frac{y^3}...

- ODE: $y'' + 4y' - 6y = 0$ - Initial conditions: $y(0) = 1$, $y'(0) = 0$ - Second task: Taylor (Maclaurin) series of $\cos x$ at $x = 0$ --- An initial value problem (IVP) is a differential equation together with the values of the unknown function and its ...

- ODE: $y'' + y = 0$ - Condition 1: $y(0) = 5$ - Condition 2: $y(\pi/4) = 3$ $$m^2 + 1 = 0 \implies m = \pm i$$ Complex roots with $\alpha = 0$, $\beta = 1$. $$y = e^{\alpha x}(A\cos\beta x + B\sin\beta x) = A\cos x + B\sin x$$ $$y(0) = A\cos 0 + B\sin 0 = ...

- Differential equation: $y'' + 4y' + 4 = 0$ - Coefficients (interpreting as homogeneous constant-coefficient ODE): coefficient of $y''$ is $1$, of $y'$ is $4$, of $y$ is $4$. Note on interpretation: As literally written, $y'' + 4y' + 4 = 0$ has a constant ...

- ODE: $y'' + 6y' + 9 = 0$ - Initial conditions: $y(0) = 2$, $y'(0) = 1$ Note: As written, the equation contains a constant term "$+9$" rather than "$+9y$". Read literally, $y'' + 6y' = -9$ is a non-homogeneous equation. However, the standard textbook probl...