Answer to Question #255269 in Differential Equations for Anne

Question

(Variable Separable). Determine the solution of the differential equation.

a.) $\alpha$ d$\beta$ + $\beta$d$\alpha$ + $\alpha$$\beta$(3d$\alpha$ + d$\beta$) =0

b.) (xy + x)dx = (x²y² + x² + y² +1)dy

c.) dx = t(1 + t²) sec²x dt

Answer 1

a)

$\alpha(1+\beta)d\beta=-\beta(1+3\alpha)d\alpha$

$\dfrac{1+\beta}{\beta}d\beta=-\dfrac{1+3\alpha}{\alpha}d\alpha$

Integrate

$\int \dfrac{1+\beta}{\beta}d\beta=-\int\dfrac{1+3\alpha}{\alpha}d\alpha$

$\ln(|\beta|)+\beta=-\ln(|\alpha|)-3\alpha+\ln C$

$\beta e^{\beta}=\dfrac{C}{\alpha e^{3\alpha}}$

$\alpha \beta e^{3\alpha+\beta}=C$

b)

$(y^2+1)(x^2+1)dy=x(y+1)dx$

$\dfrac{y^2+1}{y+1}dy=\dfrac{x}{x^2+1}dx$

Integrate

$\int\dfrac{y^2+1}{y+1}dy=\int\dfrac{x}{x^2+1}dx$

$\int\dfrac{y^2+1}{y+1}dy=\int\dfrac{y^2+2y+1-2y-2+2}{y+1}dy$

$=\int(y+1)dy-2\int dy+2\int \dfrac{dy}{y+1}$

$=\dfrac{y^2}{2}+y-2y+2\ln(|y|)+C_1$

$\dfrac{y^2}{2}-y+2\ln(|y|)=\dfrac{1}{2}\ln(x^2+1)+\ln C$

c)

$\cos^2x dx = t(1 + t^2) dt$

Integrate

$\int\cos^2x dx =\int t(1 + t^2) dt$

$\dfrac{1}{2}\int(1+\cos(2x))dx=\int(t+t^3)dt$

$\dfrac{1}{2}x+\dfrac{1}{4}\sin(2x)=\dfrac{t^2}{2}+\dfrac{t^4}{4}+C$