Sinusoids as solutions to differential equationsIt is well known that the functiont↦acos⁡(ωt)+bsin⁡(ωt)is the solution to the differential equation:x″(t)=−ω2x(t)with the initial conditions x(0)=a and x′(0)=bω. I was wondering what differential equation will be solved byf(t)=∑j=1k(ajcos⁡(ωjt)+bjsin⁡(ωjt)).It is obvious that this satisfies x(t)=∑j=1kxj(t) with xj″(t)=−ωj2xj(t) and initial conditions on xj(0) and xj′(0). I was wondering if there were any other (perhaps more natural) differential equations that are satisfied by f.

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Sinusoids as solutions to differential equationsIt is well known that the functiont↦acos⁡(ωt)+bsin⁡(ωt)is the solution to the differential equation:x″(t)=−ω2x(t)with the initial conditions x(0)=a and x′(0)=bω. I was wondering what differential equation will be solved byf(t)=∑j=1k(ajcos⁡(ωjt)+bjsin⁡(ωjt)).It is obvious that this satisfies x(t)=∑j=1kxj(t) with xj″(t)=−ωj2xj(t) and initial conditions on xj(0) and xj′(0). I was wondering if there were any other (perhaps more natural) differential equations that are satisfied by f.

Landyn Whitney · Accepted Answer

Step 1Note that cos⁡{ωt}=ejωt+e−jωt2 and sin⁡{ωt}=ejωt−e−jωt2j, so your function can be rewritten asf(t)=∑i=1k{ai−jbi2ejωit+ai+jbi2e−jωit}Step 2Obviously, we can solve this function from some differential equation with characteristic roots λ=±jωi with i=1,⋯,k. To this end, we can easily construct the characteristic functionΠi=1k(λ2+ωi2)=0Expand this equation, replace λi with f(i)(t) and we get the differential equation you want. Initial conditions are determined by values of ai and bi.

Sinusoids as solutions to differential equations It is well known that the function t \ma

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