Y'''+10y''+25y'=e^x
Just apply the usual D operator for homogeneous equations. See whether you can work it around to the solution provided here:
http://www.wolframalpha.com/input/?i=y%27%27%27%2B10y%27%27%2B25y%27%3De%5Ex
To solve the differential equation Y'''+10Y''+25Y'=e^x, we can use the method of undetermined coefficients.
Step 1: Find the complementary solution
First, we find the complementary solution by assuming that Y is of the form Yc = e^(rt), where r is a constant to be determined.
Substituting Yc into the differential equation, we get:
r^3e^(rt) + 10r^2e^(rt) + 25re^(rt) = 0
Factoring out e^(rt), we have:
e^(rt)(r^3 + 10r^2 + 25r) = 0
Since e^(rt) is never equal to zero, we can solve the equation inside the parentheses:
r^3 + 10r^2 + 25r = 0
By factoring, we find that r(r + 5)^2 = 0
Therefore, we have three roots: r = 0, r = -5, and r = -5 (multiplicity 2).
Hence, the complementary solution is Yc = C1 + C2e^(-5t) + C3te^(-5t), where C1, C2, and C3 are arbitrary constants.
Step 2: Find the particular solution
To find the particular solution Yp, we assume that Yp is of the form Yp = Ae^x, where A is a constant to be determined.
Substituting Yp into the differential equation, we get:
(A)e^x + 10(A)e^x + 25(A)e^x = e^x
Simplifying the equation, we have:
36Ae^x = e^x
Dividing both sides by e^x, we get:
36A = 1
Therefore, A = 1/36.
Hence, the particular solution is Yp = (1/36)e^x.
Step 3: Find the general solution
The general solution is the sum of the complementary solution and the particular solution.
Y = Yc + Yp
= C1 + C2e^(-5t) + C3te^(-5t) + (1/36)e^x
So, the general solution to the differential equation Y'''+10Y''+25Y'=e^x is:
Y = C1 + C2e^(-5t) + C3te^(-5t) + (1/36)e^x.
Where C1, C2, and C3 are arbitrary constants that can be determined by initial conditions or boundary conditions, if given.
To solve the given differential equation, Y'''+10Y''+25Y'=e^x, we can follow these steps:
Step 1: Find the homogeneous solution
- Set e^x equal to zero: e^x = 0
- We know that e^x is never zero, so there is no exponential part in the homogeneous solution.
- We are left with the homogeneous equation: Y'''+10Y''+25Y' = 0
Step 2: Solve the homogeneous equation
- We assume the solution to the homogeneous equation takes the form Y_h = e^(rt)
- Substituting this into the equation gives r^3e^(rt)+10r^2e^(rt)+25re^(rt) = 0
- Factoring out e^(rt), we get e^(rt)(r^3+10r^2+25r) = 0
- Simplifying further, we have r(r^2+10r+25) = 0
To find the roots of this equation, we solve the quadratic equation r^2+10r+25 = 0.
Step 3: Find the roots of the quadratic equation
- The roots of the quadratic equation r^2+10r+25 = 0 can be found using the quadratic formula.
- The quadratic formula states that for an equation of the form ax^2+bx+c=0, the roots are given by: x = (-b±√(b^2-4ac))/(2a)
- Applying this formula to our equation, we have r = (-10±√(10^2-4(1)(25)))/(2)
- Simplifying further, we have r = (-10±√(100-100))/2
- Since the discriminant is zero, we have two equal roots: r = -5
Therefore, the homogeneous solution is Y_h = C1e^(-5t) + C2te^(-5t) + C3t^2e^(-5t), where C1, C2, and C3 are constants.
Step 4: Find the particular solution
- To find the particular solution, we can use the method of undetermined coefficients.
- We assume the particular solution takes the form Y_p = Ae^x, where A is a constant to be determined.
- Substituting this into the original equation, Y'''+10Y''+25Y' = e^x, we get: A + 10A + 25A = e^x
- Simplifying, we have: 36A = e^x
- Solving for A, we have A = 1/36
Therefore, the particular solution is Y_p = (1/36)e^x.
Step 5: General solution
- The general solution to the given differential equation is the sum of the homogeneous and particular solutions.
- Therefore, the general solution is: Y = Y_h + Y_p = C1e^(-5t) + C2te^(-5t) + C3t^2e^(-5t) + (1/36)e^x
This is the solution to the given differential equation.