How can I solve this second order equation?

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Shada ahmed
Shada ahmed 2022 年 6 月 28 日
編集済み: Sam Chak 2022 年 6 月 29 日
if f=60 and k=0.105with plot the relationship between qf and fo?See the image
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Shada ahmed
Shada ahmed 2022 年 6 月 28 日
Fo^2_(2*f*k)/Qf*(pi-2*k*f)*fo-f^2=0

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回答 (2 件)

AMIT POTE
AMIT POTE 2022 年 6 月 29 日
Hi Shada,
Seecond order differential equations can be solved using dsolve() or ode45() function. You can go through the following documentation for better understanding
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Sam Chak
Sam Chak 2022 年 6 月 29 日
編集済み: Sam Chak 2022 年 6 月 29 日
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.

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Sam Chak
Sam Chak 2022 年 6 月 29 日
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')

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