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 \( v_2'(t) \).\[v_1'(t) = -\frac{4}{5} \tan 5t\]\[v_2'(t) = \frac{4}{5}\]

The given system of differential equations has two functions v1 and v2. Both are the functions of the variable t only. Observe that the system of linear differential equations has one homogenous differential equation and the other differential equation is non-homogeneous.Multiplying the differential equation cos5tv1'+sin5tv2'=0 by 5sin5t, we get, 5sin5t·cos5tv1'+5sin5t·sin5tv2'=0 which gives 5sin5t·cos5tv1'+5sin25tv2'=0. Multiplying the differential equation -5sin5tv1'+5cos5tv2'=4sec5t by cos5t, we get, -5sin5t·cos5tv1'+5cos5t·cos5tv2'=4sec5t·cos5t which gives -5sin5t·cos5tv1'+5cos25tv2'=4 because sec5t·cos5t=1 ∵ sec5t=1cos5t. Adding the differential equations 5sin5t·cos5tv1'+5sin25tv2'=0 and -5sin5t·cos5tv1'+5cos25tv2'=4 we get, 5sin25t+5cos25tv2'=0+45sin25t+cos25tv2'=45v2'=4 ∵ sin25t+cos25t=1v2'=45 Thus, we have, v2'=45.Putting v2'=45 in the differential equation 5sin5t·cos5tv1'+5sin25tv2'=0, we get, 5sin5t·cos5tv1'+5sin25tv2'=05sin5t·cos5tv1'+5sin25t45=05sin5t·cos5tv1'+=-4sin25tv1'=-4sin5t·sin5t5sin5t·cos5tv1'=-4sin5t5cos5tv1'=-45tan5t ∵ sinxcosx=tanx Therefore, v1'=-45tan5t and v2'=45 .