There are two ways to do great mathematics. The first is to be smarter than everybody else. The second way is to be stupider than everybody else — but persistent, Raoul Bott.
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
An algebraic equation typically has the following form: Expression 1 = Expression 2, where the two expressions are set equal to each other.
Separable differential equations are a special class of differential equations that can be manipulated algebraically to separate the variables, allowing us to integrate and find solutions.
A separable differential equation is any differential equation that can be written in the form $\frac{dy}{dx}=f(x)g(y)$. This form allows us to separate the variables x and y on opposite sides of the equation, facilitating integration.
Plugging our initial condition y(0) = 2 ⇒ $2^2 = 0^2+C ⇒C = 4$, hence our solution is $y^2 = x^2 + 4 ⇒ y = \sqrt{x^2 +4}$ where y(0) = 2 is a positive value, we take the positive root.
$\frac{dy}{dx}=-xy$🚀 ⇒[Separate Variables] $\frac{dy}{y}=-xdx$ ⇒[Integrate Both Sides] $\int \frac{dy}{y}=-\int xdx ⇒ ln|y| = \frac{-x^2}{2}+C ⇒ $ (where y ≠ 0)
In fact, $ln|y|+c_1 = \frac{-x^2}{2}+c_2 ⇒ ln|y| = \frac{-x^2}{2}+c$ where c = c_{2} - c_{1}. In other words, two constants can always be combined.
Exponentiate both sides to eliminate the natural logarithm: $e^{ln|y|}= e^{\frac{-x^2}{2}+C} = e^{\frac{-x^2}{2}}·e^C ⇒ |y| = ae^{\frac{-x^2}{2}} ⇒ y = ±ae^{\frac{-x^2}{2}}$ where a = e^{c}. Besides, the solution $y = ae^{\frac{-x^2}{2}}$ works for any a because $\frac{dy}{dx}=-xy = a(-x)e^{\frac{-x^2}{2}}, \frac{dy}{dx}= a(\frac{-2x}{2})e^{\frac{-x^2}{2}}= a(-x)e^{\frac{-x^2}{2}}$🚀
Therefore, we have a general solution or a “family” of solutions. If we know a initial condition, say y(0) = 3 ⇒ 3 = y = $ae^{\frac{-0^2}{2}} = a$ ⇒ y = $3e^{\frac{-x^2}{2}}$ and the ambiguity of the general solution is indeed solved.
Solve the differential equation: $\frac{dy}{dx} = x^2y+y$ ⇒[Let’s separate variables] $\frac{dy}{dx} = y(x^2+1) ⇒ \frac{1}{y}·dy = (x^2 + 1)dx ⇒[\text{Integrate Both Sides}] \int \frac{dy}{y} = \int (x^2 + 1)dx ⇒ ln|y| = \frac{x^3}{3}+x + c ⇒[\text{Exponentiate Both Sides}] e^{ln|y|} = e^{\frac{x^3}{3}+x + c} ⇒ |y| = A e^{\frac{x^3}{3}+x}$ where A = e^{C}. General solution $y = ±Ae^{\frac{x^3}{3}+x}$
$\frac{dy}{dx} = x^3·y^2$ ⇒[Let’s separate variables] $\frac{1}{y^2}·\frac{dy}{dx} =x^3⇒ \frac{1}{y^2}dy = x^3dx ⇒[\text{Integrate Both Sides}] \int \frac{dy}{y^2} = \int x^3dx ⇒ \frac{-1}{y} = \frac{x^4}{4} + C ⇒ y = \frac{1}{-(\frac{x^4}{4} + C )} = \frac{1}{-\frac{1}{4}x^4-C}$
$\frac{dy}{dx} = \frac{e^x}{y},$ with initial condition y(0) = 1, y ≠ 0.
$\frac{dy}{dx} = \frac{e^x}{y}$ ⇒[Let’s separate variables] $y·\frac{dy}{dx} = e^x ⇒ \int y·dy = \int e^x dx ⇒ \frac{y^2}{2} = e^x + C$.
y(0) = 1, $\frac{1}{2} = e^0 + C = 1 + C ⇒ C = -\frac{1}{2} ⇒ \frac{y^2}{2} = e^x -\frac{1}{2} ⇒ y^2 = 2e^2 -1 ⇒ y = ±\sqrt{2e^x-1}$. Given y(0) = 1 > 0, we choose the positive square root, so $y = \sqrt{2e^x-1}$.
Apply Initial Conditions. Apply y(0) = 1 (a). Substitute x = 0 and y = 1 into the general solution: into the previous equation, 1 = sin(1 + c), 1 + c = ^{π}⁄_{2} + 2πn. Let’s choose n = 0 for simplicity, c = ^{π}⁄_{2} -1. Therefore, our general solution is $y = sin(e^x+\frac{π}{2}-1)$.
Apply y(0) = 3 (b) into the previous equation, 3 = sin(1 + c), ⊥ (The sine function has a range of [−1, 1]). The initial condition y(0)=3 cannot be satisfied.
$\frac{dy}{dx} =[\text{Rearranging Terms}] xy + 2x -2y -4 = x(y+2) -2(y+2) = (x-2)(y+2)$.
$\frac{dy}{dx} = (x-2)(y+2) ↭[\text{Separate Variables}] \frac{dy}{y+2} = (x-2)dx ⇒[\text{ Integrate Both Sides}] \int \frac{dy}{y+2} = \int (x-2)dx ⇒ln|y+2| = \frac{1}{2}x^2 -2x + C⇒[\text{Exponentiate Both Sides}] |y + 2| = e^{\frac{1}{2}x^2 -2x + C} = e^Ce^{\frac{1}{2}x^2 -2x} ⇒ y +2 = ±Ke^{\frac{1}{2}x^2 -2x}$
However, since K is an arbitrary positive constant, ±K can be replaced with a new arbitrary constant C′, which can be any nonzero real number (positive or negative). Thus, we write: $y +2 = C’e^{\frac{1}{2}x^2 -2x} ⇒y = C’e^{\frac{1}{2}x^2 -2x} -2$. Since C’ is an arbitrary constant, we can rename it back to C for simplicity: $y = Ce^{\frac{1}{2}x^2 -2x} -2$