Proceed as in this example to find a solution of the given initial-value problem. x*y" - 2xy' + 2y = x, y(1) = 3, y(1) = -2 y(x) = Need Help? Read It
Proceed as in this example to find a solution of the given initial-value problem. x*y" - 2xy' + 2y = x, y(1) = 3, y(1) = -2 y(x) = Need Help? Read It
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![**Example 5: Using Theorem 4.8.2**
**Problem:**
Solve the initial-value problem:
\[ y'' + 4y = \sin 2x, \quad y(0) = 1, \quad y'(0) = -2. \]
**Solution:**
We solve two initial-value problems.
First, we solve \( y'' + 4y = 0, \quad y(0) = 1, \quad y'(0) = -2 \). By applying the initial conditions to the general solution \( y(x) = c_1 \cos 2x + c_2 \sin 2x \) of the homogeneous DE, we find that \( c_1 = 1 \) and \( c_2 = -1 \). Therefore, \( y_h(x) = \cos 2x - \sin 2x \).
Next, we solve \( y'' + 4y = \sin 2x, \quad y(0) = 0, \quad y'(0) = 0 \). Since the left-hand side of the differential equation is the same as the DE in Example 4, the Green’s function is the same, namely, \( G(x, t) = \frac{1}{2} \sin 2(x - t) \). With \( f(t) = \sin 2t \) we see from (10) that the solution of this second problem is \( y_p(x) = \frac{1}{2} \int_{0}^{x} \sin 2(x - t) \sin 2t \, dt \).
Finally, in view of (17) in Theorem 4.8.2, the solution of the original IVP is
\[ y(x) = y_h(x) + y_p(x) = \cos 2x - \sin 2x + \frac{1}{2} \int_{0}^{x} \sin 2(x - t) \sin 2t \, dt. \quad (19) \]
If desired, we can integrate the definite integral in (19) by using the trigonometric identity
\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \]
with \( A = 2(x](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee22cf2f-b974-4b00-a3cf-09b388e7d65d%2Fcbb34f31-7e29-4a78-98ad-694227099b3f%2For3s35m_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Example 5: Using Theorem 4.8.2**
**Problem:**
Solve the initial-value problem:
\[ y'' + 4y = \sin 2x, \quad y(0) = 1, \quad y'(0) = -2. \]
**Solution:**
We solve two initial-value problems.
First, we solve \( y'' + 4y = 0, \quad y(0) = 1, \quad y'(0) = -2 \). By applying the initial conditions to the general solution \( y(x) = c_1 \cos 2x + c_2 \sin 2x \) of the homogeneous DE, we find that \( c_1 = 1 \) and \( c_2 = -1 \). Therefore, \( y_h(x) = \cos 2x - \sin 2x \).
Next, we solve \( y'' + 4y = \sin 2x, \quad y(0) = 0, \quad y'(0) = 0 \). Since the left-hand side of the differential equation is the same as the DE in Example 4, the Green’s function is the same, namely, \( G(x, t) = \frac{1}{2} \sin 2(x - t) \). With \( f(t) = \sin 2t \) we see from (10) that the solution of this second problem is \( y_p(x) = \frac{1}{2} \int_{0}^{x} \sin 2(x - t) \sin 2t \, dt \).
Finally, in view of (17) in Theorem 4.8.2, the solution of the original IVP is
\[ y(x) = y_h(x) + y_p(x) = \cos 2x - \sin 2x + \frac{1}{2} \int_{0}^{x} \sin 2(x - t) \sin 2t \, dt. \quad (19) \]
If desired, we can integrate the definite integral in (19) by using the trigonometric identity
\[ \sin A \sin B = \frac{1}{2} [\cos (A - B) - \cos (A + B)] \]
with \( A = 2(x
![Proceed as in this example to find a solution of the given initial-value problem.
\[ x^2y'' - 2xy' + 2y = x, \, y(1) = 3, \, y'(1) = -2 \]
\[ y(x) = \text{________} \]
Need Help? [Read It]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fee22cf2f-b974-4b00-a3cf-09b388e7d65d%2Fcbb34f31-7e29-4a78-98ad-694227099b3f%2F6zrcxhp_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Proceed as in this example to find a solution of the given initial-value problem.
\[ x^2y'' - 2xy' + 2y = x, \, y(1) = 3, \, y'(1) = -2 \]
\[ y(x) = \text{________} \]
Need Help? [Read It]
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