Solve First & Second Order Differential Equation

Solve first order DE by direct integration

$$ {dy \over dx} = x^2 + 2 $$

In this case, since the right-hand side of the equation is in terms of $x$, integrate directly: \begin{align} \int {dy \over dx} dx & = \int x^2 + 2 \phantom{.} dx \\ y & = {1 \over 3}x^3 + 2x + C, \text{ where } C \text{ is an arbitrary constant} \end{align}

Solve first order DE by variable separable method

$$ {dy \over dx} = y - 1 $$

Rewrite the equation such that the variables are separated: \begin{align} \int {1 \over y - 1} \phantom{.} dy & = \int 1 \phantom{.} dx \\ \ln |y - 1| & = x + C \\ |y - 1| & = e^{x + C} \\ y - 1 & = \pm e^{x + C} \\ y - 1 & = \pm e^C . e^x \\ \\ \text{Let } & A = \pm e^C, \\ y - 1 & = A e^x \\ y & = Ae^x + 1, \text{ where } A = \pm e^C \text{ and } C \text{ is an arbitrary constant} \end{align}

Solve first order DE by provided substitution

$$ \text{Solve } {dy \over dx} = 1 + 2x - {y \over x^2} \text{ by the substitution } y = v + x^2 $$

First, differentiate the provided substitution with respect to $x$: \begin{align} {d \over dx} (y) & = {d \over dx} (v + x^2) \\ {dy \over dx} & = {dv \over dx} + 2x \end{align}

Substitute the above result and the substitution $y = v + x^2$ into the DE: \begin{align} {dv \over dx} + 2x & = 1 + 2x - {v + x^2 \over x^2} \\ {dv \over dx} & = 1 - \left( {v \over x^2} + {x^2 \over x^2} \right) \\ {dv \over dx} & = 1 - {v \over x^2} - 1 \\ {dv \over dx} & = -{v \over x^2} \end{align}

Solve the DE (by variable separable method: \begin{align} \int {1 \over v} \phantom{.} dv & = \int -{1 \over x^2} \phantom{.} dx \\ \ln |v| & = {1 \over x} + C \\ |v| & = e^{ {1 \over x} + C} \\ v & = \pm e^{ {1 \over x} + C} \\ v & = \pm e^C . e^{1 \over x} \\ \\ \text{Let } & A = \pm e^C, \\ v & = A e^{1 \over x} \end{align}

Express $y$ in terms of $x$: \begin{align} y - x^2 & = A e^{1 \over x} \\ y & = Ae^{1 \over x} + x^2, \text{ where } A = \pm e^C \text{ and } C \text{ is an arbitrary constant} \end{align}

Solve second order DE

$$ {d^2 x \over dt^2} = \sin t - t $$

Integrate twice: \begin{align} \int {d^2 x \over dt^2 } \phantom{.} dt & = \int \sin t - t \phantom{.} dt \\ {dx \over dt} & = -\cos t - {1 \over 2}t^2 + C \\ \\ \int {dx \over dt} \phantom{.} dt & = \int -\cos t - {1 \over 2}t^2 + C \phantom{.} dt \\ x & = -\sin t - {1 \over 6}t^3 + Ct + D, \text{ where } C \text{ and } D \text{ are arbitrary constants} \end{align}