Solve the following ordinary differential equations.
I) (2y+x^2+1)dy/dx+2xy-9x^2=0
ii) d^2/dx^2+3day/DX+2y=x^2
#1:
(2y+x^2+1)dy/dx + (2xy-9x^2) = 0
We can rewrite this as
(2y + x^2 + 1)dy + (2xy - 9x^2)dx = 0
Check for exactness:
Let M = 2xy - 9x^2
Let N = 2y + x^2 + 1
∂M / ∂y = ∂N / ∂x
∂(2xy - 9x^2) / ∂y = ∂(2y + x^2 + 1) / ∂x
2x = 2x
Thus it is indeed exact. To further solve,
∂F/∂x = M = 2xy - 9x^2
∫ ∂F = ∫ (2xy - 9x^2)∂x
F = (x^2)y - 3x^3 + g(y)
To get g(y), we differentiate it partially with respect to y:
∂F/∂y = ∂/∂y((x^2)y - 3x^3 + g(y)) = N
∂F/∂y = x^2 + g'(y) = N
x^2 + g'(y) = 2y + x^2 + 1
g'(y) = 2y + 1
Integrating,
g(y) = y^2 + y + C
Therefore,
F = (x^2)y - 3x^3 + y^2 + y + C
#2: This is incomplete. I can't remember solving the right side (if there is a function of x; in this case it's the x^2). I have to browse my college notes first lol:
d^2y/dx^2 + 3(dy/dx) + 2y = x^2
Rewriting, such that dy/dx = D:
(D^2 + 3D + 2)y = x^2
Left side becomes:
m^2 + 3m + 2
Factoring,
(m+1)(m+2)
m = -1 and m = -2
The general solution for something like this is,
yc = c,1*e^(m,1*x) + c,2*e^(m,2*x) + ... + c,n*e^(m,n*x)
where m is the root.
Thus, the solution for the left side, (D^2 + 3D + 2)y, is
yc = c,1*e^(-x) + c,2*e^(-2x)
I hope someone checks my work also.
hope this helps? ;u;
so far so good. The complete solution turns out to be, as I'm sure you have worked out:
y = c1 e^-x + c2 e^-2x + 1/4 (2x^2-6x+7)
I) To solve the ordinary differential equation:
(2y + x^2 + 1)dy/dx + 2xy - 9x^2 = 0
Step 1: Rearrange the terms to separate the variables:
(2y + x^2 + 1)dy = (9x^2 - 2xy)dx
Step 2: Integrate both sides:
∫(2y + x^2 + 1)dy = ∫(9x^2 - 2xy)dx
Step 3: Integrate the left side:
∫2y dy + ∫x^2 dy + ∫dy = ∫9x^2 dx - ∫2xy dx
Step 4: Simplify the integrals:
y^2 + (1/3)x^3 + y = 3x^3 - x^2y + C
Step 5: Rearrange the terms to express y:
y^2 - y + x^2y = 2x^3 - (1/3)x^3 + C
y^2 + (x^2 - 1)y = (5/3)x^3 + C
This is the solution to the given ordinary differential equation.
ii) The second differential equation is:
d^2y/dx^2 + 3dy/dx + 2y = x^2
Step 1: Rewrite the equation in the standard form:
d^2y/dx^2 + 3dy/dx + 2y - x^2 = 0
Step 2: Solve the characteristic equation:
r^2 + 3r + 2 = 0
(r + 2)(r + 1) = 0
r1 = -2, r2 = -1
Step 3: Find the complementary function:
y_c = C1e^(-2x) + C2e^(-x)
Step 4: Determine the particular integral:
Assume y_p = Ax^2 + Bx + C, since x^2 is present on the right side.
Step 5: Differentiate y_p:
dy_p/dx = 2Ax + B
d^2y_p/dx^2 = 2A
Step 6: Substitute the differential equations into the original equation:
2A + 3(2Ax + B) + 2(Ax^2 + Bx + C) - x^2 = 0
Step 7: Simplify and equate coefficients:
(2A)x^2 + (6A + 2B)x + (2A + 3B + 2C) = 0
2A = 1 (from the coefficient of x^2)
6A + 2B = 0 (from the coefficient of x)
2A + 3B + 2C = 0 (from the constant term)
Step 8: Solve the system of equations to find A, B, and C:
From 2A = 1, A = 1/2
From 6A + 2B = 0, B = -3A = -3/2
From 2A + 3B + 2C = 0: Substituting A and B gives
2(1/2) + 3(-3/2) + 2C = 0
1 - 9/2 + 2C = 0
2C = 9/2 - 1
2C = 7/2
C = 7/4
So the particular integral is y_p = (1/2)x^2 - (3/2)x + 7/4
Step 9: The general solution is the sum of the complementary function and particular integral:
y = y_c + y_p = C1e^(-2x) + C2e^(-x) + (1/2)x^2 - (3/2)x + 7/4
This is the solution to the given ordinary differential equation.
To solve ordinary differential equations, we will follow these steps:
I) (2y+x^2+1)dy/dx+2xy-9x^2=0
Step 1: Rearrange the equation to isolate the derivative term (dy/dx):
(2y+x^2+1)dy/dx = 9x^2 - 2xy
Step 2: Divide both sides by (2y+x^2+1):
dy/dx = (9x^2 - 2xy) / (2y+x^2+1)
Step 3: Separate the variables by multiplying both sides by dx and dividing by (9x^2 - 2xy):
(2y+x^2+1)/(9x^2 - 2xy) dy = dx
Step 4: Integrate both sides with respect to their respective variables:
∫ (2y+x^2+1)/(9x^2 - 2xy) dy = ∫ dx
The integral on the left side may not have a simple closed-form solution, but you can use numerical methods or software to evaluate it.
ii) d^2/dx^2 + 3da/dx + 2y = x^2
Step 1: First, we need to solve for y since this is a second-order differential equation. To do that, we can assume a particular solution of the form y = ax^2 + bx + c, where a, b, and c are constants.
Step 2: Find the first and second derivatives of y:
dy/dx = 2ax + b
d^2y/dx^2 = 2a
Step 3: Substitute y, dy/dx, and d^2y/dx^2 into the original equation:
2a + 3(2ax + b) + 2(ax^2 + bx + c) = x^2
Simplifying the equation, we get:
(2a + 2c) x^2 + (6a + 2b) x + (2a + 3b) = x^2
Step 4: Equate the coefficients of each power of x to find the values of a, b, and c.
For example, equating the coefficients of x^2:
2a + 2c = 1
And equating the coefficients of x:
6a + 2b = 0
And equating the constant coefficients:
2a + 3b = 0
Step 5: Solve the system of equations obtained in Step 4 to find the values of a, b, and c.
Now you have the solution for the given ordinary differential equations.