Generating...                               quiz07_n8

  1. Find the transformed function Y(s) for the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)+y\left(t\right)=t^2$ subject to y(0) = 0 and y'(0) = 1 .

    Y(s) = $\displaystyle {{s^3+2}\over{s^5+s^3}}$ Y(s) = $\displaystyle {{s+1}\over{s^3+2\,s}}$ Y(s) = $\displaystyle {{s^3+2}\over{s^5+2\,s^3}}$ Y(s) = $\displaystyle {{s+1}\over{s^3+s}}$

  2. Find the transformed function Y(s) for the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-2\,\left({{d}\over{d\,t}}\,y \left(t\right)\right)+y\left(t\right)=\cosh t$ subject to y(0) = −1 and y'(0) = −1 .

    Y(s) = $\displaystyle -{{s^3-s^2-2\,s+1}\over{s^4-2\,s^3-2\,s^2+2\,s+1}}$ Y(s) = $\displaystyle -{{s^3-s^2-s}\over{s^4-2\,s^3+2\,s-1}}$ Y(s) = $\displaystyle -{{s^3-s^2-s}\over{s^4-2\,s^3-2\,s^2+2\,s+1}}$ Y(s) = $\displaystyle -{{s^3-s^2-2\,s+1}\over{s^4-2\,s^3+2\,s-1}}$

  3. Find the transformed function Y(s) for the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)+2\,\left({{d}\over{d\,t}}\,y \left(t\right)\right)-2\,y\left(t\right)=e^ {- t }\,\cosh t$ subject to y(0) = 0 and y'(0) = 1 .

    Y(s) = $\displaystyle {{s^2+3\,s+1}\over{s^4+4\,s^3+2\,s^2-4\,s}}$ Y(s) = $\displaystyle {{s^2-s-1}\over{s^4-6\,s^2+4\,s}}$ Y(s) = $\displaystyle {{s^2-s+1}\over{s^4-4\,s^2+8\,s-4}}$ Y(s) = $\displaystyle {{s^2+3\,s+3}\over{s^4+4\,s^3+4\,s^2-4}}$

  4. Solve the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)+2\,\left({{d}\over{d\,t}}\,y \left(t\right)\right)+2\,y\left(t\right)=0$ subject to y(0) = 1 and y'(0) = −1 .

    $\displaystyle y\left(t\right)=e^ {- t }\,\left(\sqrt{2}\,\sinh \left(\sqrt{2}\,t \right)+\cosh \left(\sqrt{2}\,t\right)\right)$ $\displaystyle y\left(t\right)=e^ {- t }\,\left(2\,\sin t+\cos t\right)$ $\displaystyle y\left(t\right)=e^ {- t }\,\cosh \left(\sqrt{2}\,t\right)$ $\displaystyle y\left(t\right)=e^ {- t }\,\cos t$

  5. Solve the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-{{d}\over{d\,t}}\,y\left(t \right)=\cos t$ subject to y(0) = −1 and y'(0) = 1 .

    $\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}-{{7\,e^{t}}\over{4}}-{{e^ {- t }}\over{4}}+1$ $\displaystyle y\left(t\right)=-{{\sin t}\over{2}}-{{\cos t}\over{2}}-{{e^{t} }\over{2}}$ $\displaystyle y\left(t\right)=-{{\sin t}\over{2}}-{{\cos t}\over{2}}+{{3\,e^{t} }\over{2}}-2$ $\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}+{{e^{t}}\over{4}}-{{e^ {- t } }\over{4}}-1$


Department of Mathematics
Last modified: 2026-03-24