{"id":428771,"date":"2018-06-01T01:18:32","date_gmt":"2018-06-01T01:18:32","guid":{"rendered":"https:\/\/essaypaper.org\/solve-ode-matlab-assignment-help\/"},"modified":"2018-10-24T08:50:55","modified_gmt":"2018-10-24T08:50:55","slug":"solve-ode-matlab-assignment-help","status":"publish","type":"post","link":"https:\/\/www.benedictsol.com\/blogs\/solve-ode-matlab-assignment-help\/","title":{"rendered":"Solve Ode: MATLAB Assignment Help"},"content":{"rendered":"
\n\t<\/p>\n
\n

Task:<\/b><\/p>\n

Solve the following second order linear homogeneous ODE using Matlab:<\/span><\/p>\n

y<\/span><\/i>\u201d +4<\/span> y<\/span><\/i>\u2018 \u00a0\u2013 3<\/span>y<\/span><\/i> = <\/span>0,<\/span><\/p>\n

y<\/span><\/i>(0) =<\/span> 1, <\/span>y<\/span><\/i>\u2018(0) =<\/span> 0<\/span><\/p>\n

Solution:<\/b><\/p>\n

The characteristic polynomial has the following roots:<\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

The general solution of the ODE is then:<\/span>\u00a0 \u00a0\"\"<\/span><\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

We can now solve for the unknown coefficients c1, c2:<\/span>
\"\"<\/span><\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

The solution is found by substituting the correct values for c1 and c2:<\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/p>\n

Here is a better view of the solution:<\/span><\/p>\n

\"\"<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

\u00a0<\/p>\n

Now we can plot the solution:<\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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