Solution: ydy/dx=(xy? -cosxsinx)/( 1-x？ )=xy？ /( 1-x？ )-cosxsinx/( 1-x？ ).............( 1)
In order to find the solution of (1), we can first consider the equation: ydy/dx=xy? /( 1-x？ )，dy/dx=xy/( 1-x？ ),
Dy/y=xdx/( 1-x？ )=-d( 1-x？ )/[2( 1-x？ )];
Integral lny=-( 1/2)ln( 1-x? )+lnC？ =ln[C？ /√( 1-x？ )]
So y=C? /√( 1-x？ )..............(2)
Let any constant in (2) c? Replace it with the function u of x, so y=u/√( 1-x? )............(3)
Derivative of x: dy/dx=[(du/dx)/√( 1-x? )]+[ux/√( 1-x？ )? ]....................(4)
Substitute (3) and (4) into (1) to get: [u/√( 1-x? )]{[(du/dx)/√( 1-x？ )]+[ux/√( 1-x？ )? ]} = [Xu? /( 1-x？ )? ]-cosxsinx/( 1-x？ )
That is u(du/dx)/( 1-x? )+Xu? /( 1-x？ )? = Xu? /( 1-x？ )? -cosxsinx/( 1-x？ )
Then udu/dx=-cosxsinx, and udu of separation variable =-cosxsinxdx = cosxd (cosx).
U stands for integral? /2=(cos？ X)/2+C/2, so u=cosx+C, and then substitute it into (3) to get the general solution y=(cosx+C)/√( 1-x? ),
The initial condition y(0)=2 is 2= 1+C, so C= 1, so the special solution is: y=(cosx+ 1)/√( 1-x? ).
2.xydx+(2x？ +3y？ -20)dy=0, y(0)= 1 find y.
Solution: multiply both sides of the original formula by integer factor y? , too;
xy？ dx+(2x？ y？ +3y^5-20y？ )dy=0............( 1)
Because? P/？ y=4xy？ =? Q/？ X, so (1) is a fully differential equation, so the general solution is:
[0，x]∫xy？ dx+[0，y]∫(2x？ y？ +3y^5-20y？ )dy=(x？ y？ /2)+(x？ y？ /2)+(y^6)/2-5y？ =C
Do you have an X? y？ +(y^6)/2-5y？ =C
Substituting the initial conditions x=0 and y= 1 gives C= 1/2-5=-9/2.
So the special solution satisfying the initial condition is x? y？ +(y^6)/2-5y？ +9/2=0
Denominator divided by 2x? y？ +y^6- 10y？ +9=0
3.dy/dx=(-2x+y)？ -7, y(0)=0 to find y.
Solution: let u=-2x+y, then y=u+2x, so dy/dx = (dy/du) (du/dx)+d (2x)/dx = du/dx+2.
So there is du/dx+2=u? -7，du/dx=u？ -9, Du /(u? -9)=( 1/6)[ 1/(u-3)- 1/(u+3)]du = dx，
Integral (1/6) [ln (u-3)/(u+3)] = x+lnc, ln[(u-3)/(u+3)]=6x+lnC.
Substitute u=-2x+y to get the general solution: ln[(y-2x-3)/(y-2x+3)]=6x+lnC, that is, (y-2x-3)/(y-2x+3) = ce (6x).
Substituting the initial conditions x=0 and y=0 to get C=- 1, then the special solution satisfying the initial conditions is:
(y-2x-3)/(y-2x+3)=-e^(6x)