y%3De%5Ex-cosx.jpeg "(d^2-5d+6)y=e^x Cosx")
Differential Equation: (d^2-5d+6)y=e^x cosx
In this article, we will discuss the solution of the differential equation:
$(d^2-5d+6)y=e^x \cos x$
This is a second-order linear ordinary differential equation with constant coefficients. We will use the method of undetermined coefficients to find the general solution.
Homogeneous Equation
First, we will find the general solution of the homogeneous equation:
$(d^2-5d+6)y=0$
The characteristic equation of the homogeneous equation is:
$r^2-5r+6=0$
Solving for $r$, we get:
$(r-2)(r-3)=0$
So, the general solution of the homogeneous equation is:
$y_c=C_1e^{2x}+C_2e^{3x}$
where $C_1$ and $C_2$ are arbitrary constants.
Particular Solution
Now, we will find the particular solution of the non-homogeneous equation:
$(d^2-5d+6)y=e^x \cos x$
We will use the method of undetermined coefficients to find the particular solution. Assume that the particular solution has the form:
$y_p=e^x(A\cos x+B\sin x)$
Substituting this into the non-homogeneous equation, we get:
$e^x((A\cos x+B\sin x)-5(A\cos x+B\sin x)+6(A\cos x+B\sin x))=e^x\cos x$
Simplifying, we get:
$A(-4\cos x-5\sin x)+B(4\sin x-5\cos x)=\cos x$
Comparing the coefficients of $\cos x$ and $\sin x$, we get:
$-4A+5B=1$ $-5A-4B=0$
Solving for $A$ and $B$, we get:
$A=-\frac{5}{41}$ $B=\frac{4}{41}$
So, the particular solution is:
$y_p=e^x\left(-\frac{5}{41}\cos x+\frac{4}{41}\sin x\right)$
General Solution
The general solution of the differential equation is the sum of the homogeneous solution and the particular solution:
$y=y_c+y_p=C_1e^{2x}+C_2e^{3x}+e^x\left(-\frac{5}{41}\cos x+\frac{4}{41}\sin x\right)$
This is the general solution of the differential equation $(d^2-5d+6)y=e^x \cos x$.
