Generating...                               quiz01_n3

  1. Solve the separable ODE $\displaystyle \sec x\,\left({{d}\over{d\,x}}\,y\right)+\csc y=0$.

    $\displaystyle \cos y=\sin x+c$ $\displaystyle {{\sin \left(2\,y\right)}\over{4}}-{{y}\over{2}}={{\sin \left(2\,x \right)}\over{4}}+{{x}\over{2}}+c$ $\displaystyle -{{\sin \left(2\,y\right)}\over{4}}-{{y}\over{2}}=-{{\sin \left(2\, x\right)}\over{4}}+{{x}\over{2}}+c$ $\displaystyle -\sin y=c-\cos x$

  2. Solve the separable ODE $\displaystyle 2\,\left({{d}\over{d\,x}}\,y\right)+y^2-1=0$ subject to the initial condition y(0) = 4

    $\displaystyle \ln \left(y+1\right)-\ln \left(y-1\right)=x+\ln 5-\ln 3$ $\displaystyle \ln \left(y+1\right)-\ln \left(y-1\right)=x+\ln 4-\ln 2$ $\displaystyle \ln \left(y-1\right)-\ln \left(y+1\right)=x-\ln 4+\ln 2$ $\displaystyle \ln \left(y-1\right)-\ln \left(y+1\right)=x-\ln 5+\ln 3$

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

    $\displaystyle {{1}\over{y}}=-{{\ln \left(e^{x}+1\right)}\over{2}}+{{\ln \left(e ^{x}-1\right)}\over{2}}+c$ $\displaystyle -{{1}\over{y}}=-{{\ln \left(e^{x}+1\right)}\over{2}}+{{\ln \left( e^{x}-1\right)}\over{2}}+c$ $\displaystyle {{1}\over{y}}=\arctan e^{x}+c$ $\displaystyle -{{1}\over{y}}=\arctan e^{x}+c$

  4. Solve the separable ODE $\displaystyle \left(x+3\right)\,\left({{d}\over{d\,x}}\,y\right)-x-6=0$.

    $\displaystyle y=2\,\ln \left(x+5\right)+x+c$ $\displaystyle y=6\,\ln x+x+c$ $\displaystyle y=6\,\ln \left(x+1\right)+x+c$ $\displaystyle y=6\,\ln \left(x+3\right)+c$ $\displaystyle y=3\,\ln \left(x+3\right)+x+c$

  5. Solve the separable ODE $\displaystyle y\,\left({{d}\over{d\,x}}\,y\right)-2\,x^2\,\sqrt{y^2+1}=0$ subject to the initial condition y(0) = 1

    $\displaystyle {{\sqrt{y^2+1}}\over{2}}={{x^3}\over{3}}+{{\sqrt{5}}\over{2}}$ $\displaystyle {{\sqrt{y^2+1}}\over{2}}={{x^2}\over{2}}+{{\sqrt{5}}\over{2}}$ $\displaystyle {{\sqrt{y^2+1}}\over{2}}={{x^3}\over{3}}+{{1}\over{\sqrt{2}}}$ $\displaystyle {{\sqrt{y^2+1}}\over{2}}={{x^2}\over{2}}+{{1}\over{\sqrt{2}}}$


Department of Mathematics
Last modified: 2025-05-04