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Substituting these expressions into the system shown earlier yields the following system. (cos 5t)v₁' + (sin 5t)v₂' = 0 " (-5 sin 5t)v₁ + (5 cos 5t)v₂ = 4 sec 5t Solve this system for v₁ '(t) and v₂'(t). V₁ '(t) = V₂' (t) = 5 4 5 tan 5t

Substituting these expressions into the system shown earlier yields the following system. (cos 5t)v₁' + (sin 5t)v₂' = 0 " (-5 sin 5t)v₁ + (5 cos 5t)v₂ = 4 sec 5t Solve this system for v₁ '(t) and v₂'(t). V₁ '(t) = V₂' (t) = 5 4 5 tan 5t

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Please work out the step-by-step process of how you get to the solution on paper, please. The answer is given, I just want to see the work done that teaches how to get the solution. *See Image.

Substituting these expressions into the system shown earlier yields the following system. \[ (\cos 5t)v_1' + (\sin 5t)v_2' = 0 \] \[ (-5 \sin 5t)v_1' + (5 \cos 5t)v_2' = 4 \sec 5t \] Solve this system for \( v_1'(t) \) and \( v_2'(t) \). \[ v_1'(t) = -\frac{4}{5} \tan 5t \] \[ v_2'(t) = \frac{4}{5} \]
Transcribed Image Text:Substituting these expressions into the system shown earlier yields the following system. \[ (\cos 5t)v_1' + (\sin 5t)v_2' = 0 \] \[ (-5 \sin 5t)v_1' + (5 \cos 5t)v_2' = 4 \sec 5t \] Solve this system for \( v_1'(t) \) and \( v_2'(t) \). \[ v_1'(t) = -\frac{4}{5} \tan 5t \] \[ v_2'(t) = \frac{4}{5} \]
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I don't know if you know this, but when you upload your work it gets all jumbled together and makes it unreadable. That's why I asked for a worked solution on paper. Please resubmit.

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