MATH 315, Textwork Section 3.6 Solutions

• Problem 3.6.3: $L[y] = y''-2y'-3y = -3te^{-t}$.
  Homogeneous problem has solution $y_c(t) = c_1e^{-t}+c_2e^{3t}$. Now $g(t) = -3te^{-t}$, so you need $Y(t) = t(A +Bt)e^{-t}$ (extra $t$ factor is needed because $L[e^{-t}]=0$).
  Then $Y'(t) = (A+2Bt)e^{-t}-Y$ and $Y''(t) = 2Be^{-t}-2(A+2Bt)e^{-t}+Y$,
  $L[Y] = (2B-4A-8Bt)e^{-t} = -3e^{-t}$, so $B=3/8$, $A=3/16$.
  Final solution: $y(t) = c_1e^{-t}+c_2e^{3t}+3t(1+2t)e^{-t}/16$.

• Problem 3.6.4: $L[y] = y''+9y = 6+t^2e^{3t}$.
  Homogeneous problem has solution $y_c(t) = c_1cos(3t)+c_2sin(3t)$, so first consider $g_1(t) = 6$.
  You need $Y_1(t) = A$; then $Y_1'(t) = Y_1''(t) = 0$. Now $L[Y_1] = 9A=6$, so $A=2/3$.
  Next, consider $g_2(t) = t^2e^{3t}$. You need $Y_2(t) = (A+Bt+Ct^2)e^{3t}$;
  then $Y_2'(t) = (B+2Ct)e^{3t}+3(A+Bt+Ct^2)e^{3t}$ and
  $Y_2''(t) = 2Ce^{3t}+ 6(B+2Ct)e^{3t}+9(A+Bt+Ct^2)e^{3t}$.
  Now $L[Y_2] = (18Ct^2 + (18B+12C)t + 18A+6B+2C)e^{3t}=t^2e^{3t}$, so
  $18C=1$, $18B+12C=0$ and $18A+6B+2C=0$; therefore $C=1/18$, $B=-1/27$, $A=1/162$.
  Final solution: $y(t) = c_1cos(3t)+c_2sin(3t)+2/3+ (1/162-t/27+t^2/18)e^{3t}$.

• Problem 3.6.7: $L[y] = 2y''+3y'+y = t^2+3sin(t)$.
  Homogeneous problem has solution $y_c(t) = c_1e^{-t/2}+c_2e^{-t}$, so first consider $g_1(t) = t^2$.
  This has degree 2 so you need $Y_1(t) = A +Bt + Ct^2$; then $Y_1'(t) = B+2Ct$ and $Y_1''(t) = 2C$.
  Now $L[Y_1] = 4C + 3B+6Ct + A+Bt+Ct^2 = t^2$, so
  $C = 1$, $6C+B=0$, $4C+3B+A=0$; therefore $C = 1$, $B=-6$, $A=14$.
  Next, consider $g_2(t) = 3sin(t)$. You need $Y_2(t) = Asin(t)+Bcos(t)$;
  then $Y_2'(t) = Acos(t)-Bsin(t)$ and $Y_2''(t) = -Asin(t)-Bcos(t)$.
  Now $L[Y_2] = (-2A-3B+A)sin(t)+(-2B+3A+B)cos(t) = 3sin(t)$, so
  $-A-3B=3$, $-B+3A=0$; therefore $A=-3/10$, $B=-9/10$.
  Final solution: $y(t) = c_1e^{-t/2}+c_2e^{-t}+14-6t+t^2-3sin(t)/10-9cos(t)/10$.

• Problem 3.6.14: $L[y] = y''+4y = t^2+3e^t$, with $y(0)=0$, $y'(0)=2$.
  Homogeneous problem has solution $y_c(t) = c_1cos(2t)+c_2sin(2t)$, so first consider $g_1(t) = t^2$.
  You need $Y_1(t) = A +Bt + Ct^2$; then $Y_1'(t) = B+2Ct$, $Y_1''(t) = 2C$.
  Now $L[Y_1]=2C+4A+4Bt+4Ct^2=t^2$, so $C=1/4$, $B=0$.$A=-1/8$. Next, consider
  $g_2(t) = e^t$. You need $Y_2(t)=Ae^t$; then $Y_2''(t)=Ae^t$. Now $L[Y_2] = 5Ae^t=3e^t$, so $A=3/5$.
  General solution is $y(t) = c_1cos(2t)+c_2sin(2t)+t^2/4-1/8+3e^t/5$. Applying initial values,
  $y(0) = c_1-1/8+3/5=0$, so $c_1=-19/40$. $y'(0) = 2c_2+3/5=2$, so $c_2=7/10$.
  Final solution: $y(t) = -19cos(2t)/40+7sin(2t)/10+t^2/4-1/8+3e^t/5$.

• Problem 3.6.17: $L[y] = y''+4y = 3sin(2t)$, with $y(0)=2$, $y'(0)=-1$.
  Homogeneous problem has solution $y_c(t) = c_1cos(2t)+c_2sin(2t)$. so you need
  $Y(t) = t(Asin(2t)+Bcos(2t))$ (extra $t$ factor is needed because $L[sin(2t)]=0$). Then
  $Y'(t)=Asin(2t)+Bcos(2t)+2t(Acos(2t)-Bsin(2t))$,
  $Y''(t) =4(Acos(2t)-Bsin(2t))-4t(Asin(2t)+Bcos(2t))$. Now
  $L[Y]=-4Bsin(2t)+4Acos(2t)+t(0sin(2t)+0cos(2t))=3sin(2t)$, so $A=0$, $B=-3/4$.
  General solution is $y(t) = c_1cos(2t)+c_2sin(2t)-3tcos(2t)/4$.
  Applying initial values, $y(0) =c_1=2$, $y'(0) = 2c_2-3/4=-1$, so $c_2=-1/8$.
  Final solution: $y(t) = 2cos(2t)-sin(2t)/8-3tcos(2t)/4$.