Differential equations answers

Recent questions in Differential equations
Linda Seales 2022-01-18 Answered

Obtaining Differential Equations from Functions
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}^{{{2}}}-{1}\)
is a first order ODE,
\(\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{2}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}\right)}^{{{2}}}+{y}={0}\)
is a second order ODE and so on. I am having trouble to obtain a differential equation from a given function. I could find the differential equation for
\(\displaystyle{y}={e}^{{{x}}}{\left({A}{\cos{{x}}}+{B}{\sin{{x}}}\right)}\) and the steps that I followed are as follows.
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={e}^{{{x}}}{\left({A}{\cos{{x}}}+{B}{\sin{{x}}}\right)}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}\)
\(\displaystyle={y}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}\) (1)
\(\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{e}^{{{x}}}{\left(-{A}{\sin{{x}}}+{B}{\cos{{x}}}\right)}+{e}^{{{x}}}{\left(-{A}{\cos{{x}}}-{B}{\sin{{x}}}\right)}\)
\(\displaystyle={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{y}\right)}-{y}\)
using the orginal function and (1). Finally,
\(\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}-{2}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}+{2}{y}={0}\),
which is the required differential equation.
Similarly, if the function is \(y=(A \cos 2t+B \sin 2t)\), the differential equation that I get is
\(\displaystyle{\frac{{{d}^{{{2}}}{y}}}{{{\left.{d}{x}\right.}^{{{2}}}}}}+{4}{y}={0}\)
following similar steps as above.

Joanna Benson 2022-01-18 Answered

Solve: \(\displaystyle{y}{''}+{2}{y}'-{3}{y}={0}\)

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