Answer to Question #201024 in Differential Equations for Rijak

Question

Show complete solution. Use equation function.

1. Find the general solution of ada = 9x²dx- ydy.

2. Find the particular solution of y' = ycosx/1+2y² given y (x) = 1; y (0) = 1.

3. Find the general solution to the equation (cos y + 2) y' = 2x.

Answer 1

(1).$ada=9x^2dx-ydy$

Integrating both sides:

$\int ada= \int9x^2dx -\int ydy$

$\dfrac{a^2}{2}=9 \times\dfrac{x^3}{3} -\dfrac{y^2}{2} +C$

$\dfrac{a^2}{2}= 3{x^3} -\dfrac{y^2}{2} +C$

(2). $\dfrac{dy}{dx} =\dfrac{y cosx}{1+2y^2}$

On variable separation we get:

$(\dfrac{1}{y} +2y)dy=cos xdx$

Integrating both sides:

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

$log y +y^2=sinx +C$

Now it is given that:

y=1 when x=0, so put in above equation:

$log(1)+1=sin(o) +c$

$C=1$

So the solution is:

$logy+y^2=sinx +1 \Rightarrow logy+y^2-sinx -1 =0$

(3).$(cosy+2) \dfrac{dy}{dx}=2x$

$cos ydy+2dy=2xdx$

$siny+2y =x^2+c$

$siny+2y -x^2=c$