The general solution of the differential equation $\frac{d^{4}y}{d{x}^4}-2\frac{d^{3}y}{d{x}^3}+2\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y=0$ is

To determine the general solution of a linear homogeneous differential equation with constant coefficients, the initial step involves forming its characteristic equation.

Differential Equation Analysis

The given differential equation is presented as:

$\frac{d^{4}y}{d{x}^4}-2\frac{d^{3}y}{d{x}^3}+2\frac{d^{2}y}{d{x}^2}-2\frac{dy}{dx}+y=0$

This equation is identified as a fourth-order linear homogeneous differential equation because all its terms involve $y$ or its derivatives, there are no products of $y$ or its derivatives, and the coefficients are constants. The right-hand side is zero, indicating homogeneity.

Characteristic Equation Formation

To convert the differential equation into an algebraic characteristic equation, we replace each derivative $\frac{d^{n}y}{d{x}^n}$ with $m^n$. Following this substitution, the characteristic equation for the given differential equation becomes:

$m^4-2m^3+2m^2-2m+1=0$

Roots of the Characteristic Equation

The next crucial step is to find the roots of this polynomial equation. Let's denote the polynomial as $P(m)=m^4-2m^3+2m^2-2m+1$. We can try to find simple integer roots using the Rational Root Theorem or by inspection.

- **Testing for $m=1$:** Substitute $m=1$ into $P(m)$: $P(1)=(1{)}^4-2(1{)}^3+2(1{)}^2-2(1)+1=1-2+2-2+1=0$. Since $P(1)=0$, $m=1$ is a root of the equation. This implies that $(m-1)$ is a factor of the polynomial $P(m)$.

To find the remaining factor, we perform polynomial division of $m^4-2m^3+2m^2-2m+1$ by $(m-1)$:

$\frac{m^4-2m^3+2m^2-2m+1}{m-1}=m^3-m^2+m-1$

So, the characteristic equation can be expressed as:

$(m-1)(m^3-m^2+m-1)=0$

Now, we need to factorize the cubic term $m^3-m^2+m-1$. We can use grouping for this:

- Factoring the cubic term: $m^3-m^2+m-1=m^2(m-1)+1(m-1)$ $=(m^2+1)(m-1)$

Substituting this back into the characteristic equation, we get:

$(m-1)(m^2+1)(m-1)=0$

This simplifies to the fully factored form:

$(m-1{)}^2(m^2+1)=0$

Now, we can easily find the individual roots by setting each factor to zero:

- **From $(m-1{)}^2=0$:** $m-1=0\Rightarrow m=1$ This root appears twice, so it has a multiplicity of 2. Thus, we have two real and repeated roots: $m_1=1$ and $m_2=1$.
- **From $(m^2+1)=0$:** $m^2=-1\Rightarrow m=\pm \sqrt{-1}\Rightarrow m=\pm i$ These are a pair of complex conjugate roots: $m_3=i$ and $m_4=-i$.

A summary of the roots found is presented in the table below:General Solution Construction

The form of the general solution of a homogeneous linear differential equation depends on the nature of its characteristic roots:

- For real and repeated roots: If a real root $m=a$ has a multiplicity of $k$, the corresponding part of the solution is given by $(c_1+c_{2}x+\cdots +c_k{x}^{k-1})e^{ax}$. For our roots $m=1$ with multiplicity 2, the solution part is $(c_1+c_{2}x)e^{1x}=(c_1+c_{2}x)e^x$.
- For complex conjugate roots: If a pair of complex conjugate roots is $m=\alpha \pm i\beta$, the corresponding part of the solution is $e^{\alpha x}(c_3\cos (\beta x)+c_4\sin (\beta x))$. For our roots $m=\pm i$, we can write them as $0\pm i1$, so $\alpha =0$ and $\beta =1$. The solution part is $e^{0x}(c_3\cos (1x)+c_4\sin (1x))=1\cdot (c_3\cos x+c_4\sin x)=c_3\cos x+c_4\sin x$.

By combining these two parts, the general solution of the given differential equation is:

$y=(c_1+c_{2}x)e^x+c_3\cos x+c_4\sin x$

Conclusion on General Solution

The general solution for the given differential equation is found to be $y=(c_1+c_{2}x)e^x+c_3\cos x+c_4\sin x$.