Answer to Question #148361 in Differential Equations for masab dad

Question

Find Particular solution of d²y/dx² -y=sinh2x

Answer 1

Let us find the solution of the equation $\frac{d²y}{dx²} -y=\sinh2x$. Note that $\sinh 2x =\frac{e^{2x}-e^{-2x}}{2}$ .

The characteristic equation $k^2-1=0$ of the equation $\frac{d²y}{dx²} -y=0$ has the solutions $k=1$and $k=-1.$

The solution of the equation $\frac{d²y}{dx²} -y=\frac{e^{2x}-e^{-2x}}{2}$ is $y=C_1e^x+C_2e^{-x+}y_p$, where $y_p$ is a particular solution of $\frac{d²y}{dx²} -y=\frac{e^{2x}-e^{-2x}}{2}$ .

$y_p=Ae^{2x}+Be^{-2x}$

$y_p'=2Ae^{2x}-2Be^{-2x}$

$y_p''=4Ae^{2x}+4Be^{-2x}$

$4Ae^{2x}+4Be^{-2x}-(Ae^{2x}+Be^{-2x})=\frac{e^{2x}-e^{-2x}}{2}$

$3Ae^{2x}+3Be^{-2x}=\frac{1}{2}e^{2x}-\frac{1}{2}e^{-2x}$

$3A=\frac{1}{2}$ and $3B=-\frac{1}{2}$

$A=\frac{1}{6}$ and $B=-\frac{1}{6}$

Therefore, the general solution is the following:

$y=C_1e^x+C_2e^{-x}+\frac{1}{6}e^{2x}-\frac{1}{6}e^{-2x}=C_1e^x+C_2e^{-x}+\frac{1}{3}\sinh 2x$

The particular solution is $y_p=\frac{1}{3}\sinh 2x$ (for $C_1=C_2=0$).