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Example: Back into the complex root section we complete the claim that
y1 (t ) = elt cos(µt) and y2(t) = elt sin(µt)
Those were a basic set of solutions. Prove that they actually are.
Solution
Thus, to prove this we will require to take find the Wronskian for these two solutions and show that this isn't zero.
= elt cos(µt)( lelt sin(µt) + µ elt cos(µt)) - elt sin(µt)( lelt cos(µt) - µ elt sin(µt))
= µ e2lt cos2(µt) + µ e2lt sin2(µt)
= µ e2lt( cos2(µt) + sin2(µt))
= µ e2lt
Here, the exponential will never be zero and µ ≠ 0 whether it were we wouldn't have complex roots and so W ≠ 0. Thus, these two solutions are actually a fundamental set of solutions and hence the general solution in this case is. As:
y (t ) = c1elt cos (mt ) + c2eltsin (mt)
approximate value is the precise or the accurate value which is measured to the actual value.., approximation is how close the measured value is to the actual value , for example
2/2
In this case we are going to consider differential equations in the form, y ′ + p ( x ) y = q ( x ) y n Here p(x) and q(x) are continuous functions in the
236+2344+346=
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