Variation of Parameters — Question 4

Question 4 – Solution

Strategy. A basis changes the parameter functions, but the integral response with fixed initial data must remain the same.

Step 1: Compute both parameter systems. For the first basis $W=1$, so $u_1'=-g\sin t$, $u_2'=g\cos t$. The constant change-of-basis determinant is $-2$, hence $W_z=-2$. Thus $$v_1'=\tfrac 12g(\cos t-\sin t),\qquad v_2'=-\tfrac 12g(\cos t+\sin t).$$ The negative Wronskian is essential to these signs.

Step 2: Compare the reconstructed functions. For the first basis, the integrand multiplying $g(s)$ is $\sin t\cos s-\cos t\sin s=\sin(t-s)$. For the second, it is $$\tfrac 12[(\cos t+\sin t)(\cos s-\sin s) -(\cos t-\sin t)(\cos s+\sin s)],$$ which expands to the same expression. Therefore both give $$\boxed{Y_0(t)=\int_0^t\sin(t-s)g(s)\,ds.}$$ This also holds for negative $t$ using oriented integrals.

Step 3: Change the lower endpoint. Set $Y_a(t)=\int_a^t\sin(t-s)g(s)\,ds$. Then $$Y_a-Y_0=\cos t\int_0^a\sin s\,g(s)\,ds -\sin t\int_0^a\cos s\,g(s)\,ds.$$ Both coefficients are constants in $t$. Changing the lower endpoint changes a particular solution only by a homogeneous term; the data must be fitted again.

Step 4: Use a nonelementary forcing. The requested solution is $$\boxed{y=2\cos t-\sin t+\int_0^t\sin(t-s)e^{s^2}\,ds.}$$ Leibniz differentiation gives $Y_0'=\int_0^t\cos(t-s)e^{s^2}\,ds$ and $Y_0''=e^{t^2}-Y_0$. Also $Y_0(0)=Y_0'(0)=0$. Thus the equation and data hold. A definite-integral formula is a complete exact solution even without an elementary antiderivative.