Find the general solution of [latex]y^{(4)}-4 y^{\prime \prime \prime}+6 y^{\prime \prime}-4 y^{\prime}+y=0[/latex]

The characteristic equation is [latex]\lambda^{4}-4 \lambda^{3}+6 \lambda^{2}-4 \lambda+1=(\lambda-1)^{4}=0[/latex]. with the single root [latex]\lambda=1[/latex] of multiplicity 4 . Thus four linearly independent solutions are [latex]y_{1}=e^{x}, y_{2}=x e^{x}, y_{3}=x^{2} e^{x} \text { and } y_{4}=x^{3} e^{x}[/latex]. The general solution is [latex]y(x)=e^{x}\left(c_{1}+c_{2} x+c_{3} x^{2}+c_{4} x^{3}\right)[/latex].

Question 11.15.3: Find the general solution of y^(4) - 4y"' + 6y" - 4y' + y = ...

Multivariable Calculus, Linear Algebra, and Differential Equations [EXP-51866]

Find the general solution of

y^{(4)}-4 y^{\prime \prime \prime}+6 y^{\prime \prime}-4 y^{\prime}+y=0

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The characteristic equation is

$\lambda^{4}-4 \lambda^{3}+6 \lambda^{2}-4 \lambda+1=(\lambda-1)^{4}=0$ .

with the single root $\lambda=1$ of multiplicity 4 . Thus four linearly independent solutions are

$y_{1}=e^{x}, y_{2}=x e^{x}, y_{3}=x^{2} e^{x} \text { and } y_{4}=x^{3} e^{x}$ .

The general solution is

$y(x)=e^{x}\left(c_{1}+c_{2} x+c_{3} x^{2}+c_{4} x^{3}\right)$ .