# How can I solve this second order equation?

1 view (last 30 days)
Shada ahmed on 28 Jun 2022
Edited: Sam Chak on 29 Jun 2022
if f=60 and k=0.105with plot the relationship between qf and fo?See the image
Shada ahmed on 28 Jun 2022
Fo^2_(2*f*k)/Qf*(pi-2*k*f)*fo-f^2=0

AMIT POTE on 29 Jun 2022
Seecond order differential equations can be solved using dsolve() or ode45() function. You can go through the following documentation for better understanding
Sam Chak on 29 Jun 2022
It is impossible to solve the given equation because it is not a differential equation to begin with
Fo^2_(2*f*k)/Qf*(pi-2*k*f)*fo-f^2=0
There are Fo, f, k, Qf, pi, and another fo, with Fo, Qf, and fo are unknowns.
Furthermore, the underscore symbol in Fo^2_ is not intelligible from the mathematical notation perspective.
Most likely @Shada ahmed refers the "second order" to the polynomial order in the equation.

Sam Chak on 29 Jun 2022
This is not exactly the equation given, but it should give you the idea of ''visualizing' the relationship between Fo and Qf.
f = 60;
k = 0.105;
fimplicit(@(Fo, Qf) (Fo.^2) + (2*f*k)./Qf*(pi-2*k*f).*Fo - f^2, [-0.1 0.1 -4e-3 4e-3])
grid on
xlabel('Fo')
ylabel('Qf') R2020b

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