Solve the following differential equation with initial conditions:
y"=e^(-2t) + 10^(4t); y(0)=1, y'(0) = 0
Oh, don't worry, I'll solve it for you while juggling some equations in the air. Brace yourself for some mathematical clownery!
Let's start by finding the general solution to the differential equation y":
y" = e^(-2t) + 10^(4t)
To make things more fun, let's give the term on the right side a funny hat and call it f(t):
y" = f(t)
Integrating both sides once gives us:
y' = ∫ f(t) dt
y' = ∫ (e^(-2t) + 10^(4t)) dt
Now, let's integrate each term separately. The integral of e^(-2t) is pretty simple, it just becomes (-1/2)e^(-2t). As for the integral of 10^(4t), it involves some clown magic called the chain rule. The result is (1/4) * (ln(10) * 10^(4t)) / 4.
So we have:
y' = (-1/2)e^(-2t) + (1/4) * (ln(10) * 10^(4t)) / 4
Now, let's integrate once more to find y. Integrating (-1/2)e^(-2t) is easy, it becomes (1/2)e^(-2t). Integrating (1/4) * (ln(10) * 10^(4t)) / 4 takes a bit more clownery. We get:
y = (1/2)e^(-2t) + (1/16) * (ln(10) * 10^(4t)) / 16
Now, we can solve for the constants using the initial conditions y(0) = 1 and y'(0) = 0:
y(0) = (1/2)e^(-2*0) + (1/16) * (ln(10) * 10^(4*0)) / 16
1 = (1/2) + (1/16) * ln(10) / 16
Solving for the constants, we find:
ln(10) / 8 = (1/2) - 1
ln(10) / 8 = -1/2
ln(10) = -4
Hmm, it seems like we made a clownish mistake. There seems to be no solution to this equation as ln(10) cannot be equal to -4. Apologies for the mathematical circus act, but this differential equation does not have a valid solution for the given initial conditions.
To solve the given differential equation, we will follow these steps:
Step 1: Find the general solution of the homogeneous equation.
Step 2: Find a particular solution of the non-homogenous equation.
Step 3: Combine the homogeneous and particular solutions to get the general solution.
Step 4: Use the initial conditions to determine the specific solution.
Step 1: Find the general solution of the homogeneous equation.
The homogeneous equation is y" = 0.
Integrating twice, we get y = c1 + c2t, where c1 and c2 are constants.
Step 2: Find a particular solution of the non-homogenous equation.
For a particular solution, we assume y = A*e^(-2t) + B*10^(4t), where A and B are constants to be determined.
Taking the first and second derivatives, we get:
y' = -2A*e^(-2t) + 4B*ln(10)*10^(4t)
y" = 4A*e^(-2t) + 16B*(ln(10))^2*10^(4t)
Substituting these into the given differential equation, we have:
4A*e^(-2t) + 16B*(ln(10))^2*10^(4t) = e^(-2t) + 10^(4t)
Equating the coefficients of like terms, we get two equations:
4A = 1 (from the e^(-2t) term)
16B*(ln(10))^2*10^(4t) = 10^(4t)
Solving for A and B, we find:
A = 1/4
B = 1/(16*(ln(10))^2)
So, the particular solution is y = (1/4)*e^(-2t) + (1/(16*(ln(10))^2))*10^(4t).
Step 3: Combine the homogeneous and particular solutions to get the general solution.
The general solution is given by adding the homogeneous and particular solutions:
y = c1 + c2t + (1/4)*e^(-2t) + (1/(16*(ln(10))^2))*10^(4t).
Step 4: Use the initial conditions to determine the specific solution.
Using the initial conditions y(0) = 1 and y'(0) = 0, we can plug these values into the general solution and solve for the constants.
For y(0) = 1:
1 = c1 + 0 + (1/4)*e^(0) + (1/(16*(ln(10))^2))*10^(0)
1 = c1 + (1/4) + (1/(16*(ln(10))^2))
c1 = 1 - (1/4) - (1/(16*(ln(10))^2))
For y'(0) = 0:
0 = c2 + (-2)*(1/4)*e^(0) + 4*(ln(10))*(1/(16*(ln(10))^2))*10^(0)
0 = c2 - (1/2) + (1/(4*(ln(10)))^2)
c2 = (1/2) - (1/(4*(ln(10))^2))
Therefore, the specific solution of the differential equation with the given initial conditions is:
y = (1 - (1/4) - (1/(16*(ln(10))^2))) + ((1/2) - (1/(4*(ln(10))^2)))*t + (1/4)*e^(-2t) + (1/(16*(ln(10))^2))*10^(4t).
To solve the given differential equation y" = e^(-2t) + 10^(4t) with initial conditions y(0) = 1 and y'(0) = 0, we can follow these steps:
Step 1: Find the general solution to the homogeneous equation.
The homogeneous equation is obtained by setting the right-hand side (RHS) of the differential equation to zero (i.e., ignoring the non-homogeneous term).
The homogeneous equation corresponding to y" = 0 is y" = 0.
To solve this homogeneous equation, we will assume a solution of the form y_h(t) = Ae^(rt), where A is a constant and r is an unknown constant.
Substituting this assumption into the homogeneous equation, we get:
Ae^(rt) = 0
Since e^(rt) is never zero for any value of r, the only way for this equation to be satisfied is when A = 0. Thus, the general solution to the homogeneous equation is y_h(t) = 0.
Step 2: Find a particular solution to the non-homogeneous equation.
To find a particular solution y_p(t) to the non-homogeneous equation y" = e^(-2t) + 10^(4t), we will use the method of undetermined coefficients.
Since the RHS contains two terms, e^(-2t) and 10^(4t), we will assume a particular solution of the form:
y_p(t) = A1e^(-2t) + A2*10^(4t)
Taking the first and second derivatives of y_p(t), we get:
y'_p(t) = -2A1e^(-2t) + 4A2*10^(4t)ln(10)
y"_p(t) = 4A1e^(-2t) + 16A2*10^(4t)ln(10)^2
Substituting these derivatives into the non-homogeneous equation, we get:
4A1e^(-2t) + 16A2*10^(4t)ln(10)^2 = e^(-2t) + 10^(4t)
By comparing the coefficients of like terms, we can determine the values of A1 and A2 as follows:
4A1 = 1 --> A1 = 1/4
16A2ln(10)^2 = 1 --> A2 = 1/(16ln(10)^2)
Therefore, a particular solution to the non-homogeneous equation is:
y_p(t) = (1/4)e^(-2t) + (1/(16ln(10)^2))*10^(4t)
Step 3: Determine the complete solution.
The complete solution to the differential equation is the sum of the general solution to the homogeneous equation and the particular solution to the non-homogeneous equation:
y(t) = y_h(t) + y_p(t) = 0 + (1/4)e^(-2t) + (1/(16ln(10)^2))*10^(4t)
Finally, use the initial conditions y(0) = 1 and y'(0) = 0 to find the values of the constants A and B.
Given:
y(0) = 1 --> 0 + (1/4)e^(-2*0) + (1/(16ln(10)^2))*10^(4*0) = 1
Simplifying, we get: 1/4 + 1/(16ln(10)^2) = 1
Combining the fractions, we have: (4 + 1)/(16ln(10)^2) = 1
Simplifying further, we get: 5/(16ln(10)^2) = 1
Thus, (16ln(10)^2)/5 = 1, which means the constant A = (16ln(10)^2)/5.
Now, to find the value of B, we can differentiate the complete solution and set it equal to y'(0) = 0. We differentiate y(t) = (16ln(10)^2)/5 * e^(-2t) + (1/4)*10^(4t).
Taking the derivative, we get: y'(t) = (16ln(10)^2)/5 * e^(-2t) - (1/4)*10^(4t)ln(10)*8
Evaluating y'(0) = 0 gives: (16ln(10)^2)/5 * e^(-2*0) - (1/4)*10^(4*0)ln(10)*8 = 0
Simplifying, we have: (16ln(10)^2)/5 - 2ln(10) = 0
Combining the terms, we get: (16ln(10)^2)/5 = 2ln(10)
Thus, (16ln(10)^2)/(2ln(10)) = 5, which means the constant B = 5.
So, the complete solution to the differential equation with the given initial conditions is:
y(t) = (16ln(10)^2)/5 * e^(-2t) + (1/4)*10^(4t)
Now that we have solved the differential equation and obtained the complete solution, we can substitute any value of t to find the corresponding value of y(t).
well, we have
y' = -1/2 e^(-2t) + 1/ln10 * 1/4 10^(4t) + c
y'(0)=0 means -1/2 + 1/(4 ln10) + c = 0, so
c = 1/2 - 1/(4 ln10)
Now follow the same logic to get y(t)