Solve the initial value problem 2y" + 5y' - 3y = 0 y(0) =7, y'(0) = 7
Can someone please help. The answer I came up with is 35 exp(-x) - 28 exp(-3x/2), but my answer is wrong.
Hmmm. I get
y = -e^(x/2) - 7e^(-3x)
Better check your algebra.
2t^2+5t-3 = (2t-1)(t+3)
Well, it seems like the answer you came up with is not quite right. Let's see if I can lend a helping hand and solve this problem for you!
To begin, we can assume that the solution to the differential equation takes the form of y(x) = e^(rx), where r is an unknown constant. Plugging this into the equation, we get:
2(e^(rx))" + 5(e^(rx))' - 3(e^(rx)) = 0
Differentiating twice with respect to x, we have:
2(r^2)(e^(rx)) + 5r(e^(rx)) - 3(e^(rx)) = 0
Next, we can factor out e^(rx) from the equation:
e^(rx)(2r^2 + 5r - 3) = 0
Since e^(rx) doesn't equal zero, we can set the expression in parentheses equal to zero:
2r^2 + 5r - 3 = 0
Now we solve this quadratic equation. Applying the quadratic formula, we have:
r = (-5 ± sqrt(5^2 - 4(2)(-3))) / (2(2))
Simplifying that, we get:
r = (-5 ± sqrt(49)) / 4
So, we have two possible values for r:
r1 = (-5 + 7) / 4 = 1/2
r2 = (-5 - 7) / 4 = -3
Now let's find the corresponding solutions for y(x) using these values of r.
For r1 = 1/2:
y1(x) = e^(1/2 * x)
And for r2 = -3:
y2(x) = e^(-3x)
Since the differential equation is linear, the general solution y(x) can be written as a linear combination of these solutions:
y(x) = C1 * e^(1/2 * x) + C2 * e^(-3x)
To find C1 and C2, we can use the initial conditions y(0) = 7 and y'(0) = 7:
y(0) = C1 * e^(1/2 * 0) + C2 * e^(-3 * 0) = C1 + C2 = 7
y'(x) = (C1/2) * e^(1/2 * x) - 3C2 * e^(-3x)
y'(0) = (C1/2) * e^(1/2 * 0) - 3C2 * e^(-3 * 0) = (C1/2) - 3C2 = 7
Now we have a system of equations to solve:
C1 + C2 = 7
(C1/2) - 3C2 = 7
From the first equation, we can solve for C1:
C1 = 7 - C2
Substituting this into the second equation, we get:
(7 - C2)/2 - 3C2 = 7
7 - C2 - 6C2 = 14
-7C2 = 7
C2 = -1
Substituting the value of C2 back into the first equation, we find:
C1 + (-1) = 7
C1 = 8
So, the final solution to the initial value problem is:
y(x) = 8 * e^(1/2 * x) - e^(-3x)
And that's the correct answer! Hope this help brighten your day, even if I can't provide a laugh along the way!
To solve the given initial value problem, we can use the method of finding the characteristic equation and using exponential functions.
The characteristic equation for the differential equation 2y" + 5y' - 3y = 0 is obtained by assuming a solution of the form y = e^(rx):
2(r^2) + 5r - 3 = 0
To solve this quadratic equation, we can use the quadratic formula:
r = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 2, b = 5, and c = -3. Substituting these values into the formula, we get:
r = (-5 ± √(5^2 - 4(2)(-3))) / (2(2)) = (-5 ± √(25 + 24)) / 4 = (-5 ± √49) / 4 = (-5 ± 7) / 4
This gives us two possible values for r:
r1 = (-5 + 7) / 4 = 2/4 = 1/2
r2 = (-5 - 7) / 4 = -12/4 = -3
Therefore, the general solution to the differential equation is given by:
y(x) = C1 * e^(r1 * x) + C2 * e^(r2 * x)
where C1 and C2 are arbitrary constants.
We can now find the particular solution by using the initial values y(0) = 7 and y'(0) = 7.
Substituting x = 0, y = 7, and y' = 7 into the general solution, we get:
7 = C1 * e^(1/2 * 0) + C2 * e^(-3 * 0) = C1 + C2
Similarly, substituting x = 0, y = 7, and y' = 7 into the derivative of the general solution, we get:
7 = C1 * (1/2) * e^(1/2 * 0) + C2 * (-3) * e^(-3 * 0) = C1/2 - 3C2
Solving these two equations simultaneously, we can find the values of C1 and C2:
From the first equation, we have C1 = 7 - C2
Substituting C1 = 7 - C2 in the second equation, we get:
7 = (7 - C2)/2 - 3C2
Multiplying through by 2, we have:
14 = 7 - C2 - 6C2
Simplifying, we get:
0 = -7C2 - 7
Dividing through by -7, we have:
C2 = 1
Substituting C2 = 1 back into C1 = 7 - C2, we get:
C1 = 7 - 1 = 6
Therefore, the particular solution to the initial value problem is:
y(x) = 6 * e^(1/2 * x) + 1 * e^(-3 * x)
Simplifying, we have:
y(x) = 6e^(x/2) + e^(-3x)
To solve the given initial value problem, we can use the method of finding the general solution of the linear second-order homogeneous differential equation.
The given differential equation is: 2y" + 5y' - 3y = 0
To solve this, we need to find the characteristic equation associated with the differential equation. The characteristic equation is obtained by assuming a solution of the form y = e^(rx), where r is a constant.
1. Differentiating twice:
y' = re^(rx)
y" = r^2e^(rx)
2. Substituting the expressions for y, y', and y" into the differential equation:
2(r^2e^(rx)) + 5(re^(rx)) - 3(e^(rx)) = 0
3. Simplifying the equation by factoring out e^(rx):
e^(rx)(2r^2 + 5r - 3) = 0
Since e^(rx) ≠ 0 for any value of x, we can divide both sides of the equation by e^(rx):
2r^2 + 5r - 3 = 0
4. Solving the quadratic equation:
This equation can be factored as follows:
(2r - 1)(r + 3) = 0
So, we have two possibilities:
2r - 1 = 0 or r + 3 = 0
Solving for r gives:
r = 1/2 or r = -3
5. Finding the general solution:
Since we have two distinct values of r, the general solution of the differential equation is given by:
y(x) = c1e^(1/2*x) + c2e^(-3*x)
6. Applying the initial conditions:
Using the given initial conditions:
y(0) = 7 and y'(0) = 7
Substituting these into the general solution, we get two equations:
c1e^(1/2*0) + c2e^(-3*0) = 7 -> c1 + c2 = 7 [Equation 1]
(c1/2)e^(1/2*0) - 3c2e^(-3*0) = 7 -> c1/2 - 3c2 = 7
Simplifying the second equation:
c1/2 - 3c2 = 7 -> c1 - 6c2 = 14 [Equation 2]
7. Solving the system of equations:
Solving Equations 1 and 2 simultaneously, we can use any method like substitution or elimination. Let's use the elimination method to find the values of c1 and c2:
Multiply Equation 1 by 2:
2(c1 + c2) = 2(7) -> 2c1 + 2c2 = 14 [Equation 3]
Subtract Equation 2 from Equation 3:
2c1 + 2c2 - (c1 - 6c2) = 14 - 14
c1 + 8c2 = 0 -> c1 = -8c2
Substituting c1 = -8c2 into Equation 1:
-8c2 + c2 = 7
-7c2 = 7
c2 = -1
Substituting c2 = -1 into c1 = -8c2:
c1 = -8(-1) = 8
8. Writing the particular solution:
Now that we have found c1 and c2, we can substitute these values back into the general solution:
y(x) = c1e^(1/2*x) + c2e^(-3*x)
= 8e^(1/2*x) - e^(-3*x)
Therefore, the solution to the initial value problem is:
y(x) = 8e^(1/2*x) - e^(-3*x)
Note: The solution you provided, 35e^(-x) - 28e^(-3x/2), is incorrect.