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

    $\displaystyle {{\sqrt{y^2+1}}\over{4}}={{x^3}\over{3}}+{{\sqrt{5}}\over{4}}$ $\displaystyle {{\sqrt{y^2+1}}\over{4}}={{x^2}\over{2}}+{{\sqrt{5}}\over{4}}$ $\displaystyle {{\sqrt{y^2+1}}\over{4}}={{x^2}\over{2}}+{{1}\over{2^{{{3}\over{2}} }}}$ $\displaystyle {{\sqrt{y^2+1}}\over{4}}={{x^3}\over{3}}+{{1}\over{2^{{{3}\over{2}} }}}$

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

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

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

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

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

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

  5. Solve the separable ODE $\displaystyle {{d}\over{d\,x}}\,y-e^{x-4\,y}=0$.

    $\displaystyle {{e^{4\,y}}\over{4}}-e^{x}=c$ $\displaystyle -e^{4\,y}-e^{x}=c$ $\displaystyle -{{e^ {- 4\,y }}\over{4}}-e^{x}=c$ $\displaystyle {{e^{4\,y}}\over{4}}=x+c$


Department of Mathematics
Last modified: 2025-10-30