Find the value of k so that the ODE $\displaystyle \left(-2\,x\,e^{2\,y}-8\,x\,y^3\right)\,\left({{d}\over{d\,x}}\,y \right)-e^{2\,y}+k\,y^4+2\,x=0$ becomes exact.
k = 2 k = −1 k = 1 k = −2
Solve the homogeneous ODE $\displaystyle x\,y^2\,\left({{d}\over{d\,x}}\,y\right)+y^3-x^3=0$ subject to the initial condition y(1) = 2
$\displaystyle {{y^3}\over{3\,x^3}}+\ln x={{1}\over{3}}$ $\displaystyle {{x^3\,y^3}\over{3}}-{{x^6}\over{6}}={{5}\over{2}}$ $\displaystyle {{y^3}\over{3\,x^3}}+\ln x={{8}\over{3}}$ $\displaystyle {{x^3\,y^3}\over{3}}-{{x^6}\over{6}}={{1}\over{6}}$
Solve the exact ODE $\displaystyle \cos x\,\cos y\,\left({{d}\over{d\,x}}\,y\right)-\sin x\,\sin y+ \tan x=0$ .
$\displaystyle \cos x\,\sin y+\ln \csc x=c$ $\displaystyle \sin x\,\cos y+\ln \sec x=c$ $\displaystyle \cos x\,\sin y+\ln \sec x=c$ $\displaystyle \sin x\,\cos y+\ln \csc x=c$
Solve the exact ODE $\displaystyle \left(2\,x^2\,y+2\right)\,\left({{d}\over{d\,x}}\,y\right)+2\,x\,y^ 2=0$ .
$\displaystyle x^2\,y^2+2\,x=c$ $\displaystyle x^2\,y^2+2\,y+2\,x=c$ $\displaystyle x^2\,y^2+2\,y=c$ $\displaystyle x^2\,y^2=c$
Solve the homogeneous ODE $\displaystyle \left(y^3+2\,x^3\right)\,\left({{d}\over{d\,x}}\,y\right)+3\,x^2\,y= 0$ .
$\displaystyle {{x^3}\over{y^2}}-y=c$ $\displaystyle {{y^5}\over{5}}+x^3\,y^2=c$ $\displaystyle {{1}\over{3\,y^3}}+{{x^3}\over{3\,y^6}}=c$ $\displaystyle {{y^9}\over{9}}+{{x^3\,y^6}\over{3}}=c$

Department of Mathematics
Last modified: 2025-10-30