Title: First Order ODE's
Course: Demonstrating STACK (v4.3.7)

Solve the following differential equation: \[ [value]\frac{dy}{d[value]} + [value][value]+[value] = 0. \] Use the letter "c" (small case) to denote the arbitrary constant occuring in the general solution.
\(y([value])\) =
Feedback: For the equation \[ [value]\frac{dy}{d[value]} + [value][value]+[value] = 0 \] rearranging gives \[dy = \frac{(-[value][value]-[value])}[value]d[value]\] Therefore \[\int dy = \int\frac{(-[value][value]-[value])}[value]d[value] =\int(-[value]-\frac[value][value])d[value]\] Hence \[y = [value]\]

For what values of k is the following differential equation exact? \[ ( [value] ) \frac{dy}{dx} + [value] = 0.\]
k =
Feedback: As the differential equation\[ ( [value] ) \frac{d y}{dx} + [value] = 0\] is of the form \[A(x,y)\frac{dy}{dx} + B(x,y) = 0\] the condition for the differential equation to be exact is \[\frac{A(x,y)}{x} = \frac{B(x,y)}{y}\] Therefore \[[value]=[value]\] \[k=[value]\]

Solve the following differential equation: \[ [value]\frac{dy}{dx}-x(y+[value])=0\] Use the letter c to denote the arbitrary constant occurring in the general solution.
y(x) =
Feedback: For the equation \[ [value]\frac{dy}{dx} -x(y+ [value]) = 0 \] rearranging gives \[\frac{dy}{y+ [value]} = \frac{xdx}[value]\] To integrate, let \(u = [value]-x^2\) giving \(du = -2dx\) \[\int \frac{dy}{y+ [value]} = \int\frac{-du}{2[value]}\] Hence \[\ln(|y+[value]|) = [value]+c\] and \[y + [value] = [value] = [value]=c[value]\] Hence \[y = c[value]-[value]\]

Solve the following differential equation: \[ \frac{dy}{dx} - [value] -[value]x^2+[value]y^2+y^2x^2 = 0. \] Use the letter \(c\) to denote the arbitrary constant occuring in the general solution.
y(x) =
Feedback: Rearranging the differential equation gives \[\frac{dy}{dx}-[value]([value]+x^2)+y^2([value]+x^2)=0\] and therefore \[\frac{dy}{dx}=([value]-y^2)([value]+x^2),\] so that \[\frac{dy}{[value]-y^2}=([value]+x^2)dx.\] Integrating both sides yields: \[\int\frac{dy}{[value]-y^2}=\int([value]+x^2)dx\] To integrate the left hand side, rearrange \[\frac{dy}{[value]-y^2}=\frac{dy}{[value](1-\frac{y^2}[value])}\] Let \(\frac{y}[value] = \tanh(u)\) \[\frac{dy}[value]=(1-\tanh^2(u))du = (1-\frac{y^2}[value])du\] \[\frac{dy}{(1-\frac{y^2}[value])}=[value]du.\] Therefore \[\int\frac{dy}{[value]-y^2}=\frac{1}[value]\int\frac{dy}[value]=\frac{1}[value]\int[value]du=\frac{u}[value].\] Integrating the right hand side gives \[\int([value]+x^2)dx=[value]x+\frac{x^3}{3}+c.\] Therefore \[\frac{u}[value]=[value]x+\frac{x^3}{3}+c,\] so that \[u=[value]\left([value]x+\frac{x^3}{3}+c\right)\] and hence \[\tanh(u)=\tanh\left([value]\left([value]x+\frac{x^3}{3}+c\right)\right)\] Therefore \[\frac{y}[value]=\tanh\left([value]\left([value]x+\frac{x^3}{3}+c\right)\right)\] and \[y=[value]\]