$$x''+x=4te^t$$ We find the fundamental system is $$\{\cos t,\sin t\}$$ after that $$ f(t)=P_m(t)e^{λt} $$ $$ m=1;\; l=1;\; s=0;\; Q_m(t)=at+b$$ So we have to find the partial solution $$z(x)=x^sQ_m(t)e^{λt}=(at+b)e^t$$ $$z''=(at+b)e^t+2ae^t$$ We replace in the equation and we have: $$(at+b)e^t+2ae^t+(at+b)e^t=4te^t \iff at+b+a-2t=0$$ Final answer is $$y=c_1\cos t+c_2\sin t+(at+b)e^t.$$ My question is how do we find $a$ and $b$, are there any mistakes in my calculations?
solving simple system in a linear differential quasi-polynomal equation
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calculus
ordinary-differential-equations
derivatives
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0you must plugg your solution into the differential equation, the solution is given by $$x \left( t \right) =\sin \left( t \right) {\it \_C2}+\cos \left( t \right) {\it \_C1}+2\, \left( t-1 \right) {{\rm e}^{t}} $$ – 2017-01-06
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0I don't understand how do you find that $$z=2(t−1)e^t$$ – 2017-01-06
1 Answers
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You got as the last equation $$ at+b+a-2t=0. $$ Now you just compare coefficients to get the linear system $$ a-2=0\\ b+a=0 $$ which is now trivial to solve