Find the general solution to the Differential Equation 2.?

Question

2xy dy=((x^2) + (y^2))dx, y(1)=2

Answer 1

First, divide by dx:

2xyy' = x²+y²

Make the substitution w=y², w' = 2yy':

xw' = x²+w

This is now a linear differential equation. Subtracting w and dividing by x:

w' - 1/x w = x

We now multiply by an integrating factor of e^(∫-1/x dx) = e^(-ln x) = 1/x:

1/x w' - 1/x² w = 1

Integrating both sides:

w/x = x + C

Multiply by x:

w = x² + Cx

Now, since w=y²:

y² = x²+Cx
y = ±√(x²+Cx)

Now we know that y(1) = 2, so we must take the positive square root. And as for the value of C:

2 = √(1² + C*1)
4 = 1+C
C = 3

So y = √(x²+3x). And we are done.

Answer 2

We have dy/dx = (x^2 + y^2)/(2xy). Put y = ux. Then, dy/dx = u + x du/dx. Therefore,

u + x du/dx = (x^2 + u^2 x^2)/(2xux) = (1 + u^2)/2u

x du/dx = (1 + u^2)/2u - u = (1 - u^2)/(2u). So,

2u/(1- u^2) du = dx/x. Integrating both members, we get

-ln(1- u^2) = ln(x) + C, C a constant, so that 1 - u^2 = e^(-C)/x = K/x, since K = e^(-C) is a constant. It follows that

u = sqrt(1 - K/x) => y = ux = sqrt(1 - K/x) * x = sqrt(x^2 - Kx).

According to the initial condition, y =2 for x =1, so that

2 = sqrt(1 -k) => 4 = 1- k => k = -3.

So, finally, y = sqrt(x^2 + 3x). Let's check

dy/dx = (2x + 3)/(2sqrt(x^2 + 3x))

x^2 + y^2 = x^2 + x^2 + 3x = 2x^2 + 3x

(x^2 + y^2)/(2xy) = (2x + 3)/(2y) = (2x+3)/(2sqrt(x^2 + 3x)).

So, dy/dx = (x^2 + y^2)/(2xy), and our result is correct

This solution is a bit different from the previous 2. Both methods are correct, you choose the one you like best.

Answer 3

This is similar to using an integrating factor
2xyy' - y^2 = x^2
2yy' - (y^2)/x = x
Now as d(y^2)/dx = 2yy' we can use an integrating factor
= e^(int -1/x) = 1/x so
d((y^2)/x)/dx = 1
(y^2)/x = x + C
y = sqrt( x^2 + Cx)
Withi you conditions
y = sqrt(x^2 + 3x)