Help with Differential Equations???

Question

Solve the differential equation

dy/dx = (7x plus 2e^(4x) plus 5) / (5y^2 plus 3y plus 4)

For an arbitrary constant you may use any single lowercase or uppercase letter other than e, D or I (or x or y, obviously), e.g. C is a nice choice. Your answer should be an equation relating y and x.

Note. For the solution of a differential equation you should not take the absolute value of a log argument.

6) Solve the differential equation

2xy' = sqrt(25 - y^2)

satisfying the initial condition

y(1/2) = 0

Answer 1

put ys and xs on different sides

(5y^2+3y+4)dy=(7x+2e^(4x)+5)dx

then take integral of both sides.

5/3y^3+3/2y^2+4y=7/2x^2+1/2e^(4x)+5x+C

this is it, since you dont give a condition, C cant be solved for

dunno about the other one.

make it a good day

Answer 2

Solve by "separation of variables - y's on the left, x's on the right

dy/dx = (7x + 2e^(4x) + 5) / (5y² + 3y + 4)

(5y² + 3y + 4)dy = (7x + 2e^(4x) + 5)dx

∫(5y² + 3y + 4)dy = ∫(7x + 2e^(4x) + 5)dx

5y³/3 + 3y²/2 + 4y = 7x²/2 + e^(4x)/2 + 5x + C

2x y' = √(25 - y²)

2x dy/dx = √(25 - y²)

2/√(25 - y²) dy = dx/x

2∫1/√(5² - y²) dy = ∫dx/x

Let y = 5sinθ

dy = 5cosθdθ

θ = arcsin(y/5)

2∫1/√(5² - (5sinθ)²) 5cosθdθ = ∫dx/x

2/5*∫1/√(1 - sin²θ) * 5cosθdθ = ln |x| + C

2∫1/√(cos²θ) * cosθdθ = ln |x| + C

2∫1/cosθ * cosθdθ = ln |x| + C

2∫dθ = ln |x| + C

2θ = ln |x| + C

2arcsin(y/5) = ln |x| + C

y(½) = 0

2arcsin(0) = ln ½ + C

C = -ln ½

C = ln(2)

2arcsin(y/5) = ln |x| + ln(2)

2arcsin(y/5) = ln |2x|

y = 5sin(ln |2x|/2)

Note; they finally fixed the "+" problem - you can type + now.