Question #244286

Find the differential equations of the following equations by integrating factors by inspection. Show complete solution.


ydx + (x + x^3 y^2)dy = 0


1
Expert's answer
2021-10-05T17:54:46-0400

ydx + (x + x³ y²)dy = 0

Comparing with Mdx+Ndy=0 we get

M = y and N = x + x³y²

So δMδy=1,δNδx=1+3x2y2\frac{\delta{M}}{\delta{y}}=1 , \frac{\delta{N}}{\delta{x}} = 1+3x²y²

δMδyδNδx\frac{\delta{M}}{\delta{y}}≠\frac{\delta{N}}{\delta{x}}

So this is not an exact differential equation.

Given equation can be written as

y.1.dx + x(1+ x² y²)dy = 0

i.e of the form yf(xy)dx+xg(xy)dy=0

So integrating factor is 1MxNy=1xyxyx3y3=1x3y3\frac{1}{Mx-Ny}=\frac{1}{xy-xy-x³y³}=-\frac{1}{x³y³}

Multiplying both sides by integrating factor we get

yx3y3dx+x+x3y2x3y3dy=0\frac{y}{-x³y³}dx + \frac{x+x³y²}{-x³y³}dy=0

=> 1x3y2dx+(1x2y31y)dy=0-\frac{1}{x³y²}dx + (-\frac{1}{x²y³}-\frac{1}{y})dy=0

This is an exact differential equation.

So the general solution is

1x3y2dx1ydy=C-\int{\frac{1}{x³y²}dx} - \int{\frac{1}{y}}dy= C ,where CC is integration constant

=> 12x2y2lny=C\frac{1}{2x²y²}-ln|y| = C


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