Find second order linear homogeneous ODE with constant coefficients if its fundamental set of solutions is {e3t,te3t}

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Michele Tipton · Accepted Answer

According to the theory of ODE with constant coefficients, the number of elements of a fundamental set of family of solutions coincides with the order of the ODE. So, your ODE here is second order. In addition, if we take the ODE in its standard form,ay&amp;nbsp;″+by′+cy=0The equation is related toam2+bm+c=0You havem=3of twice order, so the quadratic equation above is a complete square like&amp;nbsp;(m−3)2=0&amp;nbsp;and we know that in these cases we can find another solution, using the reduction method, for instance. This latter method gave us&amp;nbsp;texp⁡(3t). Thus:(m−3)2→m2−6m+9=0→y&amp;nbsp;″−6y′+9y=0is the equation

Melinda Olson · Accepted Answer

The eigenvalues and eigenvectors for the coefficient matrix A in the linear homogeneous system:Y′=AY are λ1=3 with v1=&amp;lt;a;b&amp;gt; and λ2=3 with v2=&amp;lt;c;d&amp;gt;The fundamental form of the solution is:Y=c1e3tv1+c2te3tv2Take the second derivative, Y&#039;&#039; for the DEQYour original system will be of the form:y″−6y′+9=0to give you the double eigenvalue λ1,2=3. You can actually solve this to find the corresponding eigenvectors.

Find second order linear homogeneous ODE with constant coefficients if

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