Question #3e972

Question

1542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-08T09:19:38

$\frac{d y}{d x} = \frac{y + 1 + x^{2}}{x - x^{2} (sin y)}$ $dy/dx=(y+1+x^2)/(x-x^2(siny))$

$y d x - x d y + (1 + x^{2}) d x + x^{2} (sin y) d y = 0$ $ydx - xdy + (1+x^2)dx + x^2(siny)dy =0$

$y d x + (1 + x^{2}) d x = x d y - x^{2} (sin y) d y$ $ydx + (1+x^2)dx =xdy- x^2(siny)dy$

$d x [y + 1 + x^{2}] = d y [x - x^{2} (sin y)]$ $dx[y+1+x^2]=dy[x-x^2(siny)]$

$\frac{d y}{d x} = \frac{y + 1 + x^{2}}{x - x^{2} (sin y)}$ $dy/dx=(y+1+x^2)/(x-x^2(siny))$

Answer 2 · 2018-02-08T10:15:34

$y = x (x - cos y + C) - 1$ $y=x(x-cosy+C)-1$ .

Let us rewrite the given Diff. Eqn. as,

$(1 + x^{2}) d x + x^{2} sin y d y = x d y - y d x$ $(1+x^2)dx+x^2sinydy=xdy-ydx$ .

Dividing by $x^{2}$ $x^2$ , we have,

$(\frac{1 + x^{2}}{x^{2}}) d x + sin y d y = \frac{x d y - y d x}{x^{2}}$ $((1+x^2)/x^2)dx+sinydy=(xdy-ydx)/x^2$ .

The Right Member of the eqn. is note-worthy :

Observe that, $d (\frac{y}{x}) = \frac{x d y - y d x}{x^{2}}$ $d(y/x)=(xdy-ydx)/x^2$ .

Utilising this, we find that the given eqn. is,

$(\frac{1 + x^{2}}{x^{2}}) d x + sin y d y = d (\frac{y}{x})$ $((1+x^2)/x^2)dx+sinydy=d(y/x)$ , which looks like,

separable variable type. So, Integrating term-wise,

$\int (\frac{1 + x^{2}}{x^{2}}) d x + \int sin y d y + C = \int d (\frac{y}{x})$ $int((1+x^2)/x^2)dx+intsinydy+C=intd(y/x)$ .

$:. int(1/x^2+1)dx-cosy+C=y/x, i.e.,$

$-1/x+x-cosy+C=y/x, or,$

$y=x(x-cosy+C)-1,$ is the desired General Solution!