1.) (2x+y)dy + (2x+y+6)dx=0
2.)(5t+1)t ds + (25t-1)sdt=0
thanks. :)
To solve the given differential equations, we need to identify the type of equation and then find the integrating factor.
Let's solve them one by one:
1) (2x+y)dy + (2x+y+6)dx = 0
First, we notice that it is a linear differential equation of the form M(x,y)dx + N(x,y)dy = 0, where M(x,y) = 2x+y and N(x,y) = 2x+y+6.
To find the integrating factor (IF), we calculate the partial derivative of M with respect to y (M_y) and N with respect to x (N_x):
M_y = 1
N_x = 2
Next, divide the equation by the integrating factor, which is given by IF = e^(∫(N_x - M_y)dx):
IF = e^(∫(2 - 1)dx) = e^x
Now, multiply both sides of the equation by the integrating factor (IF):
e^x * (2x+y)dy + e^x * (2x+y+6)dx = 0
This simplifies to:
2xe^xdy + ye^xdy + e^x(2x+y+6)dx = 0
Rearranging terms, we get:
(e^x(2x+y+6)dx) + (ye^x - 2xe^x)dy = 0
The left-hand side now resembles the derivative of a product rule:
d((2x+6)e^x + ye^x) = 0
Integrating both sides, we find:
(2x+6)e^x + ye^x = C, where C is the constant of integration
This is the solution to the given differential equation.
2) (5t+1)t ds + (25t-1)sdt = 0
Similar to the first equation, it is a linear differential equation of the form M(t,s)ds + N(t,s)dt = 0, where M(t,s) = (5t+1)t and N(t,s) = (25t-1)s.
Let's calculate the partial derivatives:
M_s = 0
N_t = 25s
Since M_s and N_t are not equal, the equation is not exact. To make it exact, we need to find an integrating factor. The integrating factor (IF) is given by:
IF = e^(∫(N_t - M_s)dt) = e^(∫(25s - 0)dt) = e^(25st)
Multiply both sides of the equation by the integrating factor (IF):
e^(25st) * (5t+1)t ds + e^(25st) * (25t-1)sdt = 0
This simplifies to:
(5te^(25st) + te^(25st))ds + (25tse^(25st) - se^(25st))dt = 0
Rearranging terms, we get:
(5te^(25st) + te^(25st))ds + (25tse^(25st) - se^(25st))dt = 0
The left-hand side now resembles the derivative of a product rule:
d((5t+1)se^(25st) - e^(25st)) = 0
Integrating both sides, we find:
(5t+1)se^(25st) - e^(25st) = C, where C is the constant of integration
This is the solution to the given differential equation.