Answer to Question #199327 in Differential Equations for Anushka Bhasme

Question

1. [(e/√x ^-2√x) - y/√x] = dy/dx
2. xdy/dx +y=x³y⁶
3. dr/dx = tyanchya - r²/coax
4. y dx/dy = x-yx²siny

Answer 1

1.

$e^{ -2√x}/\sqrt x - y/\sqrt x = dy/dx$

$y'+\frac{1}{\sqrt{x}}y=e^{ -2√x}/\sqrt x$

$P=\frac{1}{\sqrt{x}},Q=e^{ -2√x}/\sqrt x$

$I.F.=e^{\int Pdx}=e^{\int \frac{1}{\sqrt{x}}dx}=e^{ 2√x}$

$ye^{ 2√x}=\int Q\cdot I.F.dx=\int \frac{e^{- 2√x}}{\sqrt{x}}e^{ 2√x}dx=\int \frac{1}{\sqrt{x}}dx=2\sqrt x+c$

2.

$y'/y^6+y^{-5}/x=x^2$

$y^{-5}=v$

$-5y'/y^6=v'$

$v'-5v/x=-5x^2$

$IF=e^{-5\int dx/x}=1/x^5$

$v\cdot IF=\int Q\cdot IF dx$

$\frac{1}{(xy)^5}=-5\int dx/x^3$

$\frac{1}{(xy)^5}=\frac{5}{2x^2}+c$

$\frac{1}{x^3y^5}=\frac{5}{2}+c=c_1$

3.

$dr/dx=-r^2/cosx$

$-dr/r^2=dx/cosx$

$1/r=ln(tanx+secx)+c$

4.

$dx/dy-x/y=-x^2siny$

$u(y)=x(y)/y,x'=yu'+u$

then:

$y^2u^2siny+yu'=0$

$u'/u^2=-ysiny$

$-1/u=ycosy-siny+c$

$-y/x=ycosy-siny+c$