1. Find a particular solution $y_p$ for the Cauchy-Euler ODE $\displaystyle x^2\,\left({{d^2}\over{d\,x^2}}\,y\right)-2\,x\,\left({{d}\over{d\, x}}\,y\right)+2\,y=x^3\,e^{x}$ .

    $\displaystyle \left(x^2-2\,x\right)\,e^{x}$ $\displaystyle \left(x^3-3\,x^2+3\,x\right)\,e^{x}$ $\displaystyle \left(x^2-x\right)\,e^{x}$ $\displaystyle x\,e^{x}$

  2. Find a particular solution $y_p$ for the nonhomogeneous linear ODE $\displaystyle {{d^2}\over{d\,t^2}}\,y-{{d}\over{d\,t}}\,y-6\,y=e^{3\,t}$ .

    $\displaystyle -e^{3\,t}$ $\displaystyle {{e^{3\,t}}\over{5}}$ $\displaystyle {{\left(5\,t-1\right)\,e^{3\,t}}\over{25}}$ $\displaystyle \left(-t-1\right)\,e^{3\,t}$

  3. Find a particular solution $y_p$ for the Cauchy-Euler ODE $\displaystyle x^2\,\left({{d^2}\over{d\,x^2}}\,y\right)-3\,x\,\left({{d}\over{d\, x}}\,y\right)+3\,y={{1}\over{x}}$ .

    $\displaystyle -{{x}\over{4}}$ $\displaystyle -{{4\,\ln x+1}\over{16\,x}}$ $\displaystyle {{1}\over{8\,x}}$ $\displaystyle -{{2\,x\,\ln x+x}\over{4}}$

  4. Solve the Cauchy-Euler ODE $\displaystyle x^2\,\left({{d^2}\over{d\,x^2}}\,y\right)-2\,x\,\left({{d}\over{d\, x}}\,y\right)+3\,y=0$ .

    $\displaystyle x^{{{3}\over{2}}}\,\left(c_{1}\,\sin \left({{\sqrt{3}\,\ln x }\over{2}}\right)+c_{2}\,\cos \left({{\sqrt{3}\,\ln x}\over{2}} \right)\right)$ $\displaystyle c_{1}\,x^{{{\sqrt{5}}\over{2}}-{{1}\over{2}}}+c_{2}\,x^{-{{\sqrt{5} }\over{2}}-{{1}\over{2}}}$ $\displaystyle {{c_{1}\,\sin \left({{\sqrt{11}\,\ln x}\over{2}}\right)+c_{2}\, \cos \left({{\sqrt{11}\,\ln x}\over{2}}\right)}\over{\sqrt{x}}}$ $\displaystyle c_{1}\,x^{{{\sqrt{13}}\over{2}}+{{3}\over{2}}}+c_{2}\,x^{{{3}\over{ 2}}-{{\sqrt{13}}\over{2}}}$

  5. Solve the Cauchy-Euler ODE $\displaystyle x^2\,\left({{d^2}\over{d\,x^2}}\,y\right)+3\,x\,\left({{d}\over{d\, x}}\,y\right)+3\,y=0$ subject to y(1) = 2 and y'(1) = $\displaystyle \sqrt{2}-2$ .

    $\displaystyle y=2\,x^3+x$ $\displaystyle y={{\sin \left(\sqrt{2}\,\ln x\right)}\over{x}}+{{2\,\cos \left( \sqrt{2}\,\ln x\right)}\over{x}}$ $\displaystyle y={{2\,\sin \left(\sqrt{2}\,\ln x\right)}\over{x}}+{{\cos \left( \sqrt{2}\,\ln x\right)}\over{x}}$ $\displaystyle y=x^3+2\,x$


Department of Mathematics
Last modified: 2026-01-23