Answer to Question #339562 in Differential Equations for Themba

Question

Find the general solution of the following differential equation using the method of

undetermined coefficients: d^y/dx^2 - dy/dx + 2y = 2cosh2x

Answer 1

At first, we solve the homogeneous equation: $g''-g'+2g=0$. The characteristic equation has the form: $\lambda^2-\lambda+2=0$. The solution is: $\lambda_1=\frac{1-i\sqrt{7}}{2}$, $\lambda_2=\frac{1+i\sqrt{7}}{2}$. Thus, $g=C_1e^{\lambda_1x}+C_2e^{\lambda_2x}$with $C_1,C_2\in{\mathbb{C}}$ is the solution of equation: $g''-g'+2g=0$. The solution can be rewritten in the following form: $g=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x)$ with $k_1,k_2\in{\mathbb{C}}$. We will try to find a solution of inhomogeneous equation in the form: $f=ae^{\alpha x}+be^{\beta x}$ with $a,b,\alpha,\beta\in{\mathbb{C}}$ . We receive: $f'=a\alpha e^{\alpha x}+b\beta e^{\beta x}$, $f''=a\alpha^2e^{\alpha x}+b\beta^2e^{\beta x}$. $f''-f'+2f=a(\alpha^2-\alpha+2)e^{\alpha x}+b(\beta^2-\beta+2)e^{\beta x}$ . Right side of equation has the form: $e^{2x}+e^{-2x}$. We put $\alpha=2$, $\beta=-2$ and receive: $f''-f'+2f=4ae^{2x}+8be^{-2x}=e^{2x}+e^{-2x}$. From the latter we get: $a=\frac14$, $b=\frac18$. A solution is $f=\frac14e^{2x}+\frac18e^{-2x}$. The general solution of the initial equation is: $y=f+g=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x)+\frac14e^{2x}+\frac18e^{-2x}$ with $k_1,k_2\in{\mathbb{C}}$.

Answer: the solution is: $y=k_1e^{\frac12x}\sin(\frac{\sqrt{7}}2x)+k_2e^{\frac12x}\cos(\frac{\sqrt{7}}2x)+\frac14e^{2x}+\frac18e^{-2x}$ with $k_1,k_2\in{\mathbb{C}}$.