The general solution of the differential equationd4ydx4-2d3ydx3-2d2ydx2-2dydx-y-0

Given DE is [D^4 2D^3+2D^2 2D+1]y=0] Auxiliary equation is m^4 2m^3+2m^2 2m+1=0 (m 1)^2(m^2+1)=0 nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; nbsp; ...

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Standard XII

Mathematics

Order of a Differential Equation

The general s...

Question

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

y = (c_{1} - c_{2} x) e^{x} + c_{3} c o s x + c_{4} s i n x

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y = (c_{1} + c_{2} x) e^{x} - c_{3} c o s x + c_{4} s i n x

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y = (c_{1} + c_{2} x) e^{x} + c_{3} c o s x + c_{4} s i n x

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y = (c_{1} + c_{2} x) e^{x} + c_{3} c o s x - c_{4} s i n x

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Solution

The correct option is C $y = (c_{1} + c_{2} x) e^{x} + c_{3} c o s x + c_{4} s i n x$
Given $D E$ is $[D^{4} - 2 D^{3} + 2 D^{2} - 2 D + 1] y = 0]$
Auxiliary equation is $m^{4} - 2 m^{3} + 2 m^{2} - 2 m + 1 = 0$
$(m - 1)^{2} (m^{2} + 1) = 0$
$m = 1, 1, 1 \pm i$
General solution is
$= (C_{1} + C_{2} x) e^{x} + C_{3} c o s x + C_{4} s i n x$

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The general solution of the differential equation d4ydx4−2d3ydx3+2d2ydx2−2dydx+y=0

The correct option is C y=(c1+c2x)ex+c3cosx+c4sinxGiven DE is [D4−2D3+2D2−2D+1]y=0] Auxiliary equation is m4−2m3+2m2−2m+1=0 (m−1)2(m2+1)=0 m=1,1,1±i General solution is =(C1+C2x)ex+C3cosx+C4sinx

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