Find the solution of the Euler equation on (0, ∞) with initial conditions x²y'' + xy' + y = 0, y(1) = 3, y’(1) = 1 y(x) =
Find the solution of the Euler equation on (0, ∞) with initial conditions x²y'' + xy' + y = 0, y(1) = 3, y’(1) = 1 y(x) =
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Solving the Euler Equation
Consider the following differential equation defined on the interval \( (0, \infty) \):
\[ x^2 y'' + x y' + y = 0 \]
where \( y(1) = 3 \) and \( y'(1) = 1 \).
To find the solution to this Euler equation, follow these steps:
1. **Substitute the general solution structure**: Assume \( y = x^r \). Substituting \( y = x^r \) into the differential equation gives:
\[ x^2 (r(r-1)x^{r-2}) + x (r x^{r-1}) + x^r = 0 \]
2. **Simplify the terms**: This simplifies to:
\[ r(r-1)x^r + r x^r + x^r = 0 \]
3. **Factor out the common term**: Factor out \( x^r \):
\[ x^r (r^2 - r + r + 1) = 0 \]
\[ x^r (r^2 + 1) = 0 \]
4. **Solve for \( r \)**: The characteristic equation is:
\[ r^2 + 1 = 0 \]
Solving for \( r \), we get:
\[ r = \pm i \]
Thus, the general solution is:
\[ y(x) = c_1 \cos(\ln(x)) + c_2 \sin(\ln(x)) \]
5. **Apply the initial conditions**:
- \( y(1) = 3 \):
\[ y(1) = c_1 \cos(0) + c_2 \sin(0) = c_1 = 3 \]
- \( y'(x) \):
\[ y'(x) = c_1 \left( -\frac{\sin(\ln(x))}{x} \right) + c_2 \left( \frac{\cos(\ln(x))}{x} \right) \]
\( y'(1) = 1 \):
\[ 1 = c_1(-\sin(0)) + c_2(\cos(0)) = c_2 \Rightarrow c_2 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d3733a5-5e9e-433b-b6d1-2cfec636672d%2Fc48e264d-1f30-4c46-b3e2-f74450eea326%2F4g6ul1r_processed.png&w=3840&q=75)
Transcribed Image Text:### Solving the Euler Equation
Consider the following differential equation defined on the interval \( (0, \infty) \):
\[ x^2 y'' + x y' + y = 0 \]
where \( y(1) = 3 \) and \( y'(1) = 1 \).
To find the solution to this Euler equation, follow these steps:
1. **Substitute the general solution structure**: Assume \( y = x^r \). Substituting \( y = x^r \) into the differential equation gives:
\[ x^2 (r(r-1)x^{r-2}) + x (r x^{r-1}) + x^r = 0 \]
2. **Simplify the terms**: This simplifies to:
\[ r(r-1)x^r + r x^r + x^r = 0 \]
3. **Factor out the common term**: Factor out \( x^r \):
\[ x^r (r^2 - r + r + 1) = 0 \]
\[ x^r (r^2 + 1) = 0 \]
4. **Solve for \( r \)**: The characteristic equation is:
\[ r^2 + 1 = 0 \]
Solving for \( r \), we get:
\[ r = \pm i \]
Thus, the general solution is:
\[ y(x) = c_1 \cos(\ln(x)) + c_2 \sin(\ln(x)) \]
5. **Apply the initial conditions**:
- \( y(1) = 3 \):
\[ y(1) = c_1 \cos(0) + c_2 \sin(0) = c_1 = 3 \]
- \( y'(x) \):
\[ y'(x) = c_1 \left( -\frac{\sin(\ln(x))}{x} \right) + c_2 \left( \frac{\cos(\ln(x))}{x} \right) \]
\( y'(1) = 1 \):
\[ 1 = c_1(-\sin(0)) + c_2(\cos(0)) = c_2 \Rightarrow c_2 =
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