Title: Second Order ODE'S
Course: Demonstrating STACK (v4.3.7)

Solve \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0,\ y(0) = [value],\ y^\prime(0) = [value]\] where \({}^\prime\) denotes differentiation with respect to \(x\).
y(x) =
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} - [value]me^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 - [value]m + [value] = 0\] which can be solved to give \(m = [value]\) The solution to the differential equation is then of the form \(y = A[value]+Bx[value]\) Given \(\ y(0) = [value]\) \[ A = [value]\] and \(\ y^\prime(0) = [value]\) \[[value] + B = [value]\] Solving for A and B gives\[ [value] \mbox{ and } [value]\] Therefore \[y = [value]\]

Solve \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} - [value] y = 0.\] Use \(A\) and \(B\) for the arbitrary constants. Warning: ensure you write multiplication signs, e.g. A*e^(...), to avoid misinterpretation.

\(y(x) =\)
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} - [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} - [value]me^{mx} - [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 - [value]m - [value] = 0\] which can be solved using the quadratic formula to give [value] The solution is then of the form \(y = A[value]+B[value]\)

Solve \[[value] \frac{d^2y}{dx^2} + [value] y = 0.\] Use A and B for the arbitrary constants. Warning: ensure you write "a*e^()", "a*exp()", "a*cos()" or "A*sin()" to avoid misinterpretation.
y(x) =
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 + [value] = 0\] which can be solved using the quadratic formula to give [value] For roots of \(m\) in the form \(a±bi\), the solution to the differential equation is of the form \(y = e^{ax}(Acos(bx)+Bsin(bx))\) Therefore \(y=Acos([value]x)+Bsin([value]x)\)

Solve \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0.\] Use A and B for the arbitrary constants. Warning: ensure you write "A*e^()", "A*exp()", "A*cos()" or "A*sin()" to avoid misinterpretation.
y(x) =
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} - [value]me^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 - [value]m + [value] = 0\] which can be solved using the quadratic formula to give [value] For roots of \(m\) in the form \(a±bi\), the solution to the differential equation is of the form \(y =e^{ax}(Acos(bx)+Bsin(bx))\) Therefore \(y = [value](Acos([value]x)+Bsin([value]x))\)

Solve \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0,\ y(0) = [value],\ y^\prime(0) = [value]\] where \({}^\prime\) denotes differentiation with respect to \(x\).
y(x) =
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} - [value] \frac{dy}{dx} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} - [value]me^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 - [value]m + [value] = 0\] which can be solved to give [value] The solution is then of the form \(y = A[value]+B[value]\) Given \(\ y(0) = [value]\) \[ A + B = [value]\] and \(\ y^\prime(0) = [value]\) \[[value] + [value] = [value]\] Solving for A and B gives\[ [value] \mbox{ and } [value]\] Therefore \[y = [value]\]

Solve \[[value] \frac{d^2y}{dx^2} + [value] y = 0,\ y(0) = [value],\ y^\prime(0) = [value]\]
where \({}^\prime\) denotes differentiation with respect to \(x\).
y(x) =
Feedback: To solve the differential equation \[[value] \frac{d^2y}{dx^2} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 + [value] = 0\] which can be solved to give [value] For roots of \(m\) in the form \(a±bi\), the solution to the differential equation is of the form \[y = [value](Acos(bx)+Bsin(bx))\] and therefore \[y= Acos([value])+Bsin([value])\] Given \(\ y(0) = [value]\) \[ A = [value]\] and \(\ y^\prime(0) = [value]\) \[B = [value]\] Therefore \[y = [value]\]

Find the most general solution to the differential equation \[[value] \frac{d^2y}{dx^2} + [value] \frac{dy}{dx} + [value] y = e^{[value]x}\]
Use the letters \({\bf A}\) and \({\bf B}\) to denote the constants in your solution.
y(x) =
Feedback: To find the complementary function, solve the differential equation \[[value] \frac{d^2y}{dx^2} + [value] \frac{dy}{dx} + [value] y = 0\] Let \(y = e^{mx}\) and hence \(\frac{dy}{dx}=me^{mx}\) and \(\frac{d^2y}{dx^2} = m^2e^{mx}\) Substituting for \(y\) gives \[[value]m^2e^{mx} + [value]me^{mx} + [value]e^{mx} = 0\] and dividing throughout by \(e^{mx}\) and [value] gives \[m^2 + [value]m + [value] = 0\] which can be solved to give [value] The solution is then of the form \[y = A[value]+Bx[value]\] To find the particular integral, let \(y = ke^{[value]x}\) and hence \(\frac{dy}{dx}=[value]ke^{[value]x}\) and \(\frac{d^2y}{dx^2} = [value]ke^{[value]x}\) giving \[[value]ke^{[value]x} + [value]ke^{[value]x} + [value]ke^{[value]x} = e^{[value]x}\] Therefore \[[value]k = 1\] and \[k= \frac{1}[value]\] Thus the solution to the differential equation is \[y = A[value]+Bx[value]+\frac{1}[value]e^[value]\]