- #1
Iacha
- 3
- 0
Hi, everyone! This is my first post here, I need an hand with this equation!
Solve the initial value problem:
\begin{equation}
\begin{cases}
u''(x)+4u(x)=\cos(2x)
\\u(0)=u'(0)=1
\end{cases}
\end{equation}
I started by solving the associated homogeneous linear equation \begin{equation} u''(x)+4u(x)=0\end{equation} by the usual method of substituting the trial solution \begin{equation} u(x)=e^{\lambda x} \end{equation}.
One obtains \begin{equation}\begin{aligned} &\lambda^2+4=0 \\ &\lambda = \pm 2i \\ &u(x)=C_1\cos(2x)+C_2\sin(2x)\end{aligned}\end{equation}
To find the particular solution I use the method of undetermined coefficients with \begin{equation}u(x)_p=x(A\cos(2x)+B\sin(2x)). \end{equation}
Differentiating and substituting back in the ODE I get \begin{equation} u(x)_p=\frac{1}{4}\cos(2x)\end{equation} which is not good. I don't think I made some mistakes in the differentiation process since I checked my calculations with Maple.
Homework Statement
Solve the initial value problem:
\begin{equation}
\begin{cases}
u''(x)+4u(x)=\cos(2x)
\\u(0)=u'(0)=1
\end{cases}
\end{equation}
The Attempt at a Solution
I started by solving the associated homogeneous linear equation \begin{equation} u''(x)+4u(x)=0\end{equation} by the usual method of substituting the trial solution \begin{equation} u(x)=e^{\lambda x} \end{equation}.
One obtains \begin{equation}\begin{aligned} &\lambda^2+4=0 \\ &\lambda = \pm 2i \\ &u(x)=C_1\cos(2x)+C_2\sin(2x)\end{aligned}\end{equation}
To find the particular solution I use the method of undetermined coefficients with \begin{equation}u(x)_p=x(A\cos(2x)+B\sin(2x)). \end{equation}
Differentiating and substituting back in the ODE I get \begin{equation} u(x)_p=\frac{1}{4}\cos(2x)\end{equation} which is not good. I don't think I made some mistakes in the differentiation process since I checked my calculations with Maple.