Answer to Question #214230 in Differential Equations for SID

Question

Let x

α

y

β be an integrating factor of x(4ydx + 2xdy) + y

3

(3ydx + 5xdy) = 0. Find α, β and

the solution of given differential equation.

Answer 1

Question

Let x$^\alpha$y$^\beta$ be an integrating factor of x (4ydx+2xdy)+y³(3ydx+5xdy)=0. Find $\alpha and \beta$ and solution of the given differential equation.

Solution.

The equation can be written as;

4xydx +2x²dy +3y⁴dx+5xy³dy=0

(4xy+3y⁴)dx+(2x²+5xy³)dy=0

Is an equation of the form ;

P(x,y)dx+Q(x,y)dy=0

$\frac {dP}{dy}$=4x+12y³

$\frac {dQ}{dx}$ =4x+5y³

$\frac {dP}{dy}\not = \frac {dQ}{dx}$ ;the equation is not exact.

We multiply the equation with the interesting factor for it to be exact.

x$^\alpha$y$^\beta$ (4xy+3y⁴)dx+x$^\alpha y^\beta$(2x²+5xy³)dy=0

(4x$^{1+\alpha}y^{1+\beta}+3x^\alpha y^{4+\beta}$ )dx +(2$x^{2+\alpha}y^\beta+5x^{1+\alpha}y^{3+\beta})$ dy=0

For the equation to be exact;

$\frac {dP}{dy}=\frac {dQ}{dx}$

$4(1+\beta)x^{1+\alpha}y^\beta+3(4+\beta)x^\alpha y^{3+\beta}=2(2+\alpha)x^{1+\alpha}y^\beta+5(1+\alpha)x^\alpha y^{3+\beta}$

Equate like terms together;

4(1+$\beta)=2(2+\alpha)$ ;$\alpha=2\beta$

3($4+\beta)=5(1+\alpha)$

Both give ;

$\beta=1,\alpha=2$

The integrating factor $x^\alpha y^\beta=x^2y$y

P(x,y)dx+Q(x,y)dy=0

Becomes,

(4x³y²+3x²y⁵)dx+(2x⁴y+5x³y⁴)dy=0

Integration of either Pdx or Qdy gives;

x⁴y²+x³y⁵=C

Answers;

$\alpha=2$

$\beta=1$

Solution;

x⁴y²+x³y⁵=C