Partial derivatives mathematics?
Find the general solution of
6y(dy/dx) = (7x^2) - 3x + 9
Q1) 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.
Your equation is not a solution of the differential equation.
Hint: the differential equation is separable. Rearrange it in the form f(y) dy = g(x) dx and then integrate to integral of dy = g(x) dx
ANSWER : y(x) =
Q2) Find the particular solution, given that when x = 1, y = 3.
ANSWER: y(x) =
Comments
There is no need for brackets around single terms as you have here.
Find the general solution by separating the variables then integrating:
6y(dy / dx) = 7x² - 3x + 9
y dy = (7x² / 6 - x / 2 + 3 / 2) dx
∫ y dy = ∫ (7x² / 6 - x / 2 + 3 / 2) dx
y² / 2 = 7x³ / 18 - x² / 4 + 3x / 2 + C
y² = 7x³ / 9 - x² / 2 + 3x + C
y = ±√(7x³ / 9 - x² / 2 + 3x + C)
y = ±√[(14x³ - 9x² + 54x + C) / 18]
y = ±√(14x³ - 9x² + 54x + C) / √18
y = ±√(14x³ - 9x² + 54x + C) / (3√2)
Find the particular solution by solving for the constant:
When x = 1, y = 3
±√(59 + C) / (3√2) = 3
±√(59 + C) = 9√2
59 + C = 162
C = 103
y = ±√(14x³ - 9x² + 54x + 103) / (3√2)
Only the positive root works hence,
y = √(14x³ - 9x² + 54x + 103) / (3√2)
6y(dy/dx) = (7x^2) - 3x + 9
[6y]dy = [(7x^2) - 3x + 9]dx
Integrate on bothsides
6y^2/ 2 = 7x^3/ 3 - 3x^2/2 + 9x + c1
3y^2 = 7x^3/3 - 3x^2 /2 + 9x + c1
y^2 = 7x^3 - x^2/2 + 3x/2 + c------------------------(1) where c1/3 = c
Now When x = 1 and y = 3 put them in Eq.(1)
(3)^2 = 7(1)^3 - (1)^2/2 + 3(1)/2 + c
9 = 7 - 1/2 + 3/2 + c
9 = 8 + c
c = 9 -8
c = 1
Now put in Eq.(1)
The solution is
y^2 = 7x^3 - x^2/2 + 3x/2 + 1----------------------------(2)
6y*dy = (7x^2-3x+9)dx
Integrate
3y^2 = (7/3)x^3 -(3/2)x^2 +9x + c
y^2 = (7/9)x^3 -(1/2)x^2 + 3x +C.........................Ans
Hold on