Can you help me to solve these differential equations?

Question

Dy/dx + xy = xy2 when x=0, y=2

(1-x2)*Dy/dx – xy = xy2

(Dy/dx)^2 = (x+1)/(x+2)2

Ln (Dy/dx) = x+y; when x=0, y=0

Answer 1

the idea behind these is to write dy/dx in terms of some product of f(y) and g(x), so like dy/dx = e^x*e^y for the last one....
once you do that you can separate dy and dx and get an integral like dy/f(y) = dxg(x), then integrate both sides to get an expression H(y) = L(x) + C. then you can solve for y in terms of x. if you have boundary conditions (like x=0, y=0) then you can use those to solve for the added constant from integration.
so it's three steps really:
1) get dy/dx in terms of some product of a function of y by a function of x. dy/dx = f(x)*g(x)
2) rewrite it as dy/f(y) = dx*g(x) , then integrate.
3) using the integrated form solve for y in terms of x.
ex.
1) isolating dy/dx
so ln(dy/dx) = x+y
e^(ln(dy/dx))= e^(x+y) = e^x*e^y
dy/dx = e^x*e^y
2) rewriting and integrating
dy/e^y, or dy *e^(-y) = e^x dx
integrate against respective variables.
-e^(-y) = e^x + C.
3)solving for y in terms of x
plugging in x=0 and y = 0 you get
-e^(-0) = e^0 + C
or -1 = 1 + C, so C = -2.
solving for y you get y = ln(2-e^-x)

Answer 2

All of your differential equations are separable differential equations.

(1)
Rewrite the differential equations as
dy/dx = x y^2 - x y

Factor out the x on the right hand side
dy/dx = x ( y^2 - y)

Seperate variables
dy / (y^2 - y) = x dx

Integrate to get (use partial fractions on the left hand side)
ln( abs( (y-1) / y ) ) = (1/2) x^2 + C

Substitute in x=0 and y=2 to solve for C
ln(1/2) = C

Thus we have
ln( abs( (y-1) / y ) ) = (1/2) x^2 + ln(1/2)

In this case we can solve for y explicitly by taking the exponential of both sides

(y-1) / y = (1/2) e^( (1/2) x^2 )

solving for y we get
y = - 2 / [ e^( (1/2) x^2 ) - 2 ]

(2)
Just like number 1

Rewrite
(1 - x^2) dy/dx = x (y^2 + y)

Seperate variables
dy / (y^2 + y) = x dx / (1 - x^2)

Integrate (use partial fractions)
ln( abs( y / (y+1) ) ) = (-1/2) ln( abs( 1 - x^2 ) ) + C

Thus
y / (y+1) = C / sqrt( abs( 1 - x^2 ) )

Solving for y we get
y = 1 / [ C sqrt( abs( 1 - x^2 ) ) - 1 ]

(3)
There are 2 differential equations to solve
dy/dx = sqrt(x+1) / (x+2)

and

dy/dx = -sqrt(x+1) / (x+2)

Just integrate ( use a change of variables, let u = sqrt(x+1) ) to get the 2 answers.

y = 2 sqrt(x+1) - 2 arctan( sqrt(x+1) ) + C
and
y = -2 sqrt(x+1) + 2 arctan( sqrt(x+1) ) + C

(4)
Take the exponential of both sides to get
dy/dx = e^(x+y)

This is a separable equation (e^(x+y) = e^x e^y). Separate variables to get
e^(-y) dy = e^x dx

Integrate
- e^(-y) = e^x + C

Substitute in x=0 and y=0
-1 = 1 + C
C = 2

Solve for y
y = - ln( 2 - e^x )

Good Luck!

Answer 3

dy/dx= x(y^2-y)
dy/(y^2-y) =x*dx so using partial fraction
ln I(y-1)/y I =x^2/2+C
At x= 0 -ln 2 =C
(1-x^2)(dy/dx) =x(1+y^2)
dy/(1+y^2) =x dx/(1-x^2) integrating
Arctan(y) =-1/2 ln I1-x^2 I +C
I don´t know where the square is so i leave it to you . in any case if you take sqrt there is a+-

dy/dx =e^(x+y) so dy/dx = e^x*e^y and

e^-y*dy = e^x*dx so -e^-y = e^x +C and at 0
-1=1+C so C= -2
e^x+e^-y-2=0

Answer 4

(x - 3) dy/dx = a million - y dy / (a million - y) = dx / (x - 3) -ln|a million - y| = ln|x - 3| + c . . . the place c is a few consistent. replace x = 4 and y = 0.5 -ln0.5 = ln1 + c ln2 = 0 + c c = ln2 -ln|a million - y| = ln|x - 3| + ln2 ln| a million/(a million-y) | = ln2|x - 3| a million/(a million - y) = 2x - 6 a million - y = a million / (2x - 6) y = a million - a million/(2x - 6) y = (2x-7)/(2x-6)