Help please...

some diff equation prbolems...

1.) Transform 2xdy - y(x+1)dx + 6y^3dx = 0 into a Bernoulli Equation.

2.) The equation of the tangent line of the one-family of curve x^2 + y^2 = 25 at
P(4, 3) is_______

3.) The complete solution of 2x^3 dy = y (y^2 + 3x^2) is_______

help please...

See:

http://www.jiskha.com/display.cgi?id=1260801721

http://www.jiskha.com/display.cgi?id=1260799498

Sure, I can help you with these differential equation problems. Let's tackle each one step by step:

1.) To transform the given equation 2xdy - y(x+1)dx + 6y^3dx = 0 into a Bernoulli Equation, we need to make it in the form of dy/dx + P(x)y = Q(x)y^n. Here's how to do it:

Let's divide the entire equation by dx:
2xdy/dx - y(x+1) + 6y^3 = 0

Now, let's divide through by y^3:
2(y^-2)dy/dx - (1/y^3)(x+1) + 6 = 0

Next, let's substitute u = y^-2:
2d(u)/dx - (1/y^3)(x+1) + 6 = 0

Now, let's multiply the entire equation by dx:
2d(u) - (1/y^3)(x+1)dx + 6dx = 0

Finally, let's group the terms:
2d(u) + (6 - (1/y^3)(x+1))dx = 0

We can see that this equation is now in the form dy/dx + P(x)y = Q(x)y^n, which is a Bernoulli Equation.

2.) to find the equation of the tangent line of the family of curves x^2 + y^2 = 25 at point P(4, 3), we can find the derivative dy/dx and evaluate it at point P. Here's how to do it:

Differentiating both sides of the equation x^2 + y^2 = 25:
2x + 2y(dy/dx) = 0

Now, let's solve for dy/dx:
dy/dx = -x/y

Substituting the coordinates of point P(4, 3):
dy/dx = -4/3

So, the slope of the tangent line is -4/3.

Using the point-slope form of the equation of a line (y-y1 = m(x-x1)), with the slope -4/3 and point P(4, 3), we have:
y - 3 = (-4/3)(x - 4)

Simplifying the equation:
3y - 9 = -4x + 16

Rearranging the equation:
4x + 3y = 25

Therefore, the equation of the tangent line is 4x + 3y = 25.

3.) To find the complete solution of 2x^3 dy = y(y^2 + 3x^2), we can separate variables and integrate. Here's how to do it:

Dividing both sides of the equation by y(y^2 + 3x^2):
(2x^3/y)(dy) = dx

Now, integrating both sides with respect to their respective variables:
∫(2x^3/y)dy = ∫dx

Integrating the left side:
∫(2x^3/y)dy = 2∫(x^3/y)dy

Simplifying the integral:
2∫(x^3/y)dy = 2∫(x^3/y^(n-1))y^(n-1)dy

Using the power rule for integration:
2∫(x^3/y^(n-1))y^(n-1)dy = 2(x^3/y^n)

Integrating the right side:
∫dx = x

Combining the integrals:
2(x^3/y^n) = x + C

Rearranging the equation:
2x^3 = y^n(x + C)

Taking the nth root of both sides:
(2x^3)^(1/n) = (y + C)^(1/n)

Finally,
y = (2x^3)^(1/n) - C

So, the complete solution to the differential equation is y = (2x^3)^(1/n) - C.

I hope this helps! Let me know if you have any more questions.

Of course! I'm here to help you with your differential equation problems. Let's tackle each question step by step:

1.) To transform the given equation into a Bernoulli Equation, we need to make a substitution of the form y = u^m, where m is a constant to be determined. Let's proceed with the steps:

a) Divide the equation by dx:
(2x * dy - y(x + 1) * dx + 6y^3 * dx) / dx = 0

b) Simplify:
2x * dy - y(x + 1) + 6y^3 = 0

c) Multiply by dx:
2x * dy - y(x + 1)dx + 6y^3dx = 0

Now, we can substitute y = u^m and dy = m * u^(m-1) * du into the equation:

2x * m * u^(m-1) * du - (u^m)(x + 1)dx + 6(u^m)^3 * dx = 0

2xmdu - (x + 1)u^m dx + 6u^(3m)dx = 0

d) Group similar terms and rearrange:
2xmdu + (6u^(3m) - (x + 1)u^m)dx = 0

Now, equating the terms multiplied by du to zero and the terms multiplied by dx to a function, we can compare their exponents and coefficients.

2xmdu = 0 => m = 0

(6u^(3m) - (x + 1)u^m)dx = f(x)

Note: f(x) represents a function of x only.

2.) To find the equation of the tangent line to the family of curves, we'll differentiate the given equation implicitly the equation x^2 + y^2 = 25 with respect to x.

Differentiating both sides, we get:
2x + 2yy' = 0

Now, we substitute the point P(4, 3) to find the slope of the tangent line at that specific point. Thus, we have:
2(4) + 2(3)y' = 0
8 + 6y' = 0
6y' = -8
y' = -8/6
y' = -4/3

Using the point-slope form of a line, the equation of the tangent line is:
y - y1 = m(x - x1)
y - 3 = (-4/3)(x - 4)
y - 3 = -4/3x + 16/3
y = -4/3x + 16/3 + 9/3
y = -4/3x + 25/3

3.) To find the complete solution of the given differential equation, we'll separate and integrate the variables. Let's proceed with the steps:

a) Separate the variables:
2x^3dy = y(y^2 + 3x^2)dx

b) Divide both sides by (y(y^2 + 3x^2)):
(2x^3dy) / (y(y^2 + 3x^2)) = dx

c) Integrate both sides with respect to their respective variables:
∫ (2x^3dy) / (y(y^2 + 3x^2)) = ∫ dx

On the left side, the substitution u = y^2 + 3x^2 will be useful. Applying the substitution, we have:

1/2 ∫ (2x^3 / u) * (du / 2x) = ∫ dx

Simplifying further,

1/2 ∫ (x^2 / u) du = ∫ dx

1/2 ∫ (x^2 / (y^2 + 3x^2)) d(y^2 + 3x^2) = ∫ dx

Letting C be the constant of integration, we get:

1/2 * ln|y^2 + 3x^2| = x + C

Simplifying further,

ln|y^2 + 3x^2| = 2x + 2C

Exponentiating both sides,

|y^2 + 3x^2| = e^(2x + 2C)

Since the absolute value can be positive or negative, we'll consider both cases:

y^2 + 3x^2 = e^(2x + 2C) (Case 1)

or

y^2 + 3x^2 = -e^(2x + 2C) (Case 2)

These are the complete solutions to the differential equation 2x^3 dy = y (y^2 + 3x^2).

I hope this helps! Let me know if you have any further questions.