Differential Equations?

1)

dy/dx = ( 2x^2y + y^3 ) / ( 2x^3 - xy^2 )

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

Let u = y/x , then y = ux , and dy/dx = u + x du/dx

u + x du/dx = u ( 2x^2 + u^2x^2 ) / ( 2x^2 - u^2x^2 )

all the x^2's on the right cancel out,

x du/dx = u( 2 + u^2) / ( 2 - u^2 ) - u

getting a common denomonator on the right

x du/dx = ( 2u + u^3 - u (2-u^2) ) / ( 2- u^2)

x du/dx = 2u^3 / ( 2- u^2 )

this is now seperable.

du ( 2- u^2 ) / 2u^3 = dx/x

break up the numerator over the denomonator on the left

du u^(-3) - (1/2) du / u = dx / x

Integrate both sides.

-(1/2) u^(-2) - (1/2) Ln | u | = Ln | x | + C

Multiplby everything by 2.

- u^(-2) - Ln | u | = Ln | x^2 | + C

replace u = y / x

- x^2 / y^2 - Ln | y / x | = Ln | x^2 | + C

C = - x^2 / y^2 - Ln | y/x | - Ln | x^2 |

use the log property to combine the Ln's together

C = - x^2 / y^2 - Ln | yx |

Multiply by -1 to match the answer above

C = x^2 / y^2+ Ln | yx |

2) dy/dx = y^2 / [ xy - x√ ( y^2 - x^2 ) ]

do the same trick, u = y/x , y = ux and dy/dx = u + x du/dx

u + x du/dx = u^2x^2 / [ x^2u - x √ ( x^2u^2 - x^2 ) ]

again, all the x's on the right hand side cancel out

x du/dx = u^2 / [ u - √ ( u^2 - 1 ) ] - u

getting a common denomonator and reducing you get

x du/dx = √ ( u^2 - 1 ) / [ u - √ ( u^2 - 1 ) ]

this is now seperable.

du [ u - √ ( u^2 - 1 ) ] / √ ( u^2 - 1 ) = dx / x

seperate the numerator over the denomonator on the left hand side.

u du / √ ( u^2 - 1 ) - du = dx / x

now integrate everything, in the very left hand term you'll have to use a substitution, let z = u^2 -1

then dz = 2u du, so ( 1/2 ) dz = u du

( 1/2 ) dz / z^(-1/2) - du = dx / x

√ ( u^2 - 1 ) - u = Ln | x | + C

now replace u = y/x

√ ( (y / x )^2 - 1 ) - y/x = Ln | x | + C

√ ( y^2 - x^2 ) / x - y / x = Ln | x | + C

Multiply everything by x.

√ ( y^2 - x^2 ) - y = x Ln | x | + Cx

This isn't your answer, and not sure if it could be simplified to equal it, but I'm fairly certain I haven't made any mistakes in this second one.

When your in DE you have to get used to the back of the book having answers which someone simplified in some god awful way which you cannot match. Just trust yourself =D.

dy/dx = (x^2)(8 + y) First, we separate the variables by making use of multiplying by making use of dx and dividing by making use of (8 + y): dy/(8 + y) = (x^2)dx combine the two facets, remembering the left will use a organic log and the splendid might have a persevering with: ln(8 + y) = (a million/3)x^3 + C develop e to the flexibility of the two facets: 8 + y = Ae^((a million/3)x^3) Subtract 8 and we've the final type: y = Ae^((a million/3)x^3) - 8 Now, we've the ingredient (0, 3), so plug those in to locate a and the particular answer: 3 = Ae^((a million/3)(0)^3) - 8 resolve for A: 11 = Ae^((a million/3)(0)) 11 = Ae^(0) A = 11 So, the particular answer is: y = 11e^((a million/3)x^3) - 8