Solve the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-{{d}\over{d\,t}}\,y\left(t \right)=\sinh t$ subject to y(0) = 0 and y'(0) = 1 .
$\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}+{{e^{t}}\over{4}}-{{e^ {- t } }\over{4}}$ $\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}-{{5\,e^{t}}\over{4}}+{{e^ {- t }}\over{4}}+1$ $\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}-{{7\,e^{t}}\over{4}}-{{e^ {- t }}\over{4}}+2$ $\displaystyle y\left(t\right)={{t\,e^{t}}\over{2}}+{{3\,e^{t}}\over{4}}+{{e^ {- t }}\over{4}}-1$
Find the transformed function Y(s) for the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-{{d}\over{d\,t}}\,y\left(t \right)-y\left(t\right)=\sinh t$ subject to y(0) = 0 and y'(0) = 1 .
Y(s) = $\displaystyle {{s^2}\over{s^4-s^3+s^2+s-2}}$ Y(s) = $\displaystyle {{s^2+s+1}\over{s^4-s^3-s-1}}$ Y(s) = $\displaystyle {{s^2}\over{s^4-s^3-2\,s^2+s+1}}$ Y(s) = $\displaystyle {{s^2+s+1}\over{s^4-s^3+3\,s^2-s+2}}$
Find the transformed function Y(s) for the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-2\,\left({{d}\over{d\,t}}\,y \left(t\right)\right)+y\left(t\right)=e^{2\,t}\,\cosh t$ subject to y(0) = 0 and y'(0) = 0 .
Y(s) = $\displaystyle {{s-2}\over{s^4-6\,s^3+12\,s^2-10\,s+3}}$ Y(s) = $\displaystyle {{1}\over{s^4-6\,s^3+14\,s^2-14\,s+5}}$ Y(s) = $\displaystyle {{s+2}\over{s^4+2\,s^3-4\,s^2-2\,s+3}}$ Y(s) = $\displaystyle {{1}\over{s^4+2\,s^3-2\,s^2-6\,s+5}}$
Solve the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-{{d}\over{d\,t}}\,y\left(t \right)-y\left(t\right)=e^{t}$ subject to y(0) = −1 and y'(0) = 1 .
$\displaystyle y\left(t\right)=e^{{{t}\over{2}}}\,\left(\sqrt{3}\,\sin \left({{ \sqrt{3}\,t}\over{2}}\right)-\cos \left({{\sqrt{3}\,t}\over{2}} \right)\right)$ $\displaystyle y\left(t\right)=e^{{{t}\over{2}}}\,\left({{3\,\sinh \left({{\sqrt... ...2}}\right)}\over{\sqrt{5}}}-\cosh \left({{\sqrt{5}\,t }\over{2}}\right)\right)$ $\displaystyle y\left(t\right)={{4\,e^{{{t}\over{2}}}\,\sinh \left({{\sqrt{5}\,t }\over{2}}\right)}\over{\sqrt{5}}}-e^{t}$ $\displaystyle y\left(t\right)=e^{{{t}\over{2}}}\,\left({{2\,\sin \left({{\sqrt{... ...t)}\over{\sqrt{3}}}-2\,\cos \left({{\sqrt{3}\,t }\over{2}}\right)\right)+e^{t}$
Solve the second-order linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y\left(t\right)-{{d}\over{d\,t}}\,y\left(t \right)-2\,y\left(t\right)=0$ subject to y(0) = 1 and y'(0) = −1 .
$\displaystyle y\left(t\right)=e^{{{t}\over{2}}}\,\left({{\sinh \left({{\sqrt{5}... ...2}}\right)}\over{\sqrt{5}}}+\cosh \left({{\sqrt{5}\,t}\over{ 2}}\right)\right)$ $\displaystyle y\left(t\right)=e^ {- t }$ $\displaystyle y\left(t\right)={{2\,e^{2\,t}}\over{3}}+{{e^ {- t }}\over{3}}$ $\displaystyle y\left(t\right)=e^{{{t}\over{2}}}\,\left(\cosh \left({{\sqrt{5}\,... ...ight)-{{3\,\sinh \left({{\sqrt{5}\,t}\over{2}}\right) }\over{\sqrt{5}}}\right)$

Department of Mathematics
Last modified: 2025-06-02