Quiz

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

$\displaystyle x^2\,y^2=c$
$\displaystyle x^2\,y^2+2\,y+2\,x=c$
$\displaystyle x^2\,y^2+2\,y=c$
$\displaystyle x^2\,y^2+2\,x=c$

Solve the exact ODE $\displaystyle \left(x+1\right)\,\left({{d}\over{d\,x}}\,y\right)+y=0$ subject to the initial condition y(1) = 1

xy + y + x = 3
xy + y = 2
xy = 2
xy + x = 3

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

$\displaystyle y={{c}\over{x}}-{{x}\over{2}}$
$\displaystyle y=x\,\ln x+c\,x$
$\displaystyle y=c\,x-x\,\ln x$
$\displaystyle y={{x}\over{2}}+{{c}\over{x}}$

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

$\displaystyle {{y^5}\over{5}}+x^3\,y^2=c$
$\displaystyle {{x^3}\over{y^2}}-y=c$
$\displaystyle {{y^9}\over{9}}+{{x^3\,y^6}\over{3}}=c$
$\displaystyle {{1}\over{3\,y^3}}+{{x^3}\over{3\,y^6}}=c$

Find the value of k so that the ODE $\displaystyle \left(8\,x\,y^3-2\,x\,e^{2\,y}\right)\,\left({{d}\over{d\,x}}\,y \right)-e^{2\,y}+k\,y^4+2\,x=0$ becomes exact.

k = −2
k = −1
k = 1
k = 2