Hi Please can you help? Really sorry for repeating the question but I wrote it incorrect before.
Find the particular solution of the differential equation from
dy/dx=2/27(x-3)�ã(x^2-6x+23) / (y) (y>0.
for which y = 2 when x = 1, and then give this particular solution in explicit form?
thank you in advance for any help
regards Claire
Sorry that the question posted contains a illegible character, which probably explains why the last question was not answered.
dy/dx=2/27(x-3)�ã(x^2-6x+23) / (y) (y>0.
I have put the illegible character in bold. It is not legible for users who use the encoding Western ISO-8859-1. You may use a different encoding, and the character may appear perfectly legible. Please describe the character if you cannot use Western ISO-8859-1 encoding in your browser. (square-root, etc.)
Also, insert sufficient parentheses around the numerator AND the denominator to avoid ambiguity. For example it is not clear if (x-3) goes with 2 or 27, and the (square-root?) term goes with 2 or (x-3).
Ideally, there should be a pair of parentheses before and after EVERY division operator.
Sure, I can help you with that!
To find the particular solution of the given differential equation, we can use the method of separation of variables.
First, let's separate the variables by multiplying both sides of the equation by y:
y * dy/dx = 2/27(x-3)�ã(x^2-6x+23)
Now, let's integrate both sides with respect to x:
∫ y dy = ∫ (2/27(x-3)�ã(x^2-6x+23)) dx
Integrating both sides gives us:
(y^2)/2 + C₁ = ∫ (2/27(x-3)�ã(x^2-6x+23)) dx
To simplify the integration on the right side, we can expand the denominator:
= (y^2)/2 + C₁ = ∫ (2/27(x-3)�ã(x^2-6x+9+14)) dx
= (y^2)/2 + C₁ = ∫ (2/27(x-3)�ã((x-3)^2 +14)) dx
Now, we can make a substitution to simplify the integral. Let u = x-3, then du = dx:
= (y^2)/2 + C₁ = ∫ (2/27u�ã(u^2+14)) du
To continue integrating, we can expand the square root:
= (y^2)/2 + C₁ = ∫ (2/27u�ãu^2+14u) du
= (y^2)/2 + C₁ = (2/27) ∫ u�ãu^2 du + (2/27) ∫ u�ã14u du
Now, integrate each term separately:
= (y^2)/2 + C₁ = (2/27) ∫ u^3/2 du + (2/27) ∫ 14u^(3/2) du
= (y^2)/2 + C₁ = (2/27) * (2/5)u^(5/2) + (2/27) * 14 * (2/5)u^(5/2)
= (y^2)/2 + C₁ = (4/135)u^(5/2) + (56/135)u^(5/2)
= (y^2)/2 + C₁ = (60/135)u^(5/2)
= (y^2)/2 + C₁ = (4/9)u^(5/2)
Now, substitute back u = x-3:
(y^2)/2 + C₁ = (4/9)(x-3)^(5/2)
Next, we can use the initial condition y = 2 when x = 1 to find the value of the constant C₁. Plugging these values into the equation, we have:
(2^2)/2 + C₁ = (4/9)(1-3)^(5/2)
4/2 + C₁ = (4/9)(-2)^(5/2)
2 + C₁ = (4/9)(-2)^(5/2)
Now, solve for C₁:
C₁ = (4/9)(-2)^(5/2) - 2
Now, substitute the value of C₁ back into our previous equation:
(y^2)/2 + (4/9)(-2)^(5/2) - 2 = (4/9)(x-3)^(5/2)
This is the explicit form of the particular solution to the given differential equation.