Math,Solve the differential equation Question

Question

Math,Solve the differential equation:y'' + 4y' + 5y = e^(-2x) * sin(x) + x^2Requirements: Show full steps using method of undetermined coefficients.Final answer + general solution.

Answer

y'' + 4y' + 5y = e^(-2x) * sin(x) + x^2y'' + 4y' + 5y = e^(-2x) ( sin(x) + e^(2x)* x^2 )assume a solution of the form

y = e^{rx}

the characteristic equation: 1*r² + 4*r + 5 = 0using the quadratic formula to find the roots (

r

):r = [-4 ± √( 4^2 - 4*1*5 )] / 2 = -2 ± isince the roots are complex (

\alpha = -2, \beta = 1

), general solution to homogeneous differential equation: y(hom) (x) = e^(-2x)*[c₁*cos(x) + c₂*sin(x)] the non-homogeneous term consists of two distinct parts: e^(-2x) * sin(x) and x^2part1 e^(-2x) * sin(x): it is in form : e^(-2x) ( a sin(x) + b cos(x) ) to eliminate this duplication, we must multiply by

x

:y1 = xe^(-2x) ( a sin(x) + b cos(x) )so, y1' = e^(-2x) (-2ax + bx + a) cos(x) + ( -ax - 2bx + b) sin(x) y1'' = e^(-2x) (3ax - 4bx - 4a) cos(x) + ( 4ax + 3bx - 4b) sin(x)plugging y1 , y1' and y1'' into left side (y'' + 4y' + 5y) and simplifying yields:e^(-2x) * 2bcos(x) - 2asin(x) = e^(-2x) sin(x)equating the coefficients of cos(x) and sin(x)2b = 0 or b = 0 and -2a = 1 or a = -1/2thus, y1 = -1/2 * xe^(-2x) cos(x)part2 x^2: it is in standard quadratic polynomial formy2 = cx^2 + dx + ey2' = 2cx + dy2" = 2csubstitute these into the differential equation:2c + 4(2cx + d) + 5(cx^2 + dx + e) = x^2(5c) x^2 + (8c + 5d)x + (2c + 4d + 5e) = x^2equating both sides :5c = 1 or c = 1/58c + 5d = 0 or d = -8/252c + 4d + 5e = 0 or e = 22/125thus, y2 = (1/5)x^2 - (8/25)x + 22/125hence the general solution is y = y(hom) (x) + y1 + y2y = e^(-2x)*[c₁*cos(x) + c₂*sin(x)] - 1/2 * xe^(-2x) cos(x) + (1/5)x^2 - (8/25)x + 22/125or y = e^(-2x)*[ (c₁ - 1/2 *x)*cos(x) + c₂*sin(x)] + (1/5)x^2 - (8/25)x +22/125

Answer

step 1: solve the homogeneous equation
y'' + 4y' + 5y = 0
characteristic equation:
r^2 + 4r + 5 = 0
solve using the quadratic formula:
r = [-4 ± sqrt(16 - 20)] / 2
r = [-4 ± sqrt(-4)] / 2
r = -2 ± i
since the roots are complex (α = -2, β = 1), the homogeneous solution is:
y_h = e^(-2x) [ c1 cos(x) + c2 sin(x) ]
step 2: set up the particular solution
since the right-hand side is a sum of two different functions, apply the principle of superposition and solve each part separately:
part a: e^(-2x) sin(x)
part b: x^2
step 3: check for resonance (part a)
normally, the guess for e^(-2x) sin(x) would be:
y_p1 = e^(-2x) (a cos x + b sin x)
but &mdash; compare this to y_h. since y_h already contains e^(-2x)cos(x) and e^(-2x)sin(x), this guess would duplicate the homogeneous solution.
this is resonance. following the modification rule, multiply by x:
corrected guess:
y_p1 = x e^(-2x) (a cos x + b sin x)
step 4: solve part a using substitution
to simplify differentiation, let:
y_p1 = e^(-2x) g(x), where g(x) = a x cos x + b x sin x
for any function of the form y = e^(-2x) g(x):
y' = e^(-2x) (g' - 2g)
y'' = e^(-2x) (g'' - 4g' + 4g)
substitute into y'' + 4y' + 5y:
e^(-2x)(g'' - 4g' + 4g) + 4e^(-2x)(g' - 2g) + 5e^(-2x)g = e^(-2x) sin x
simplify (e^(-2x) cancels):
g'' - 4g' + 4g + 4g' - 8g + 5g = sin x
g'' + g = sin x
this reduced equation is much easier to work with.
now compute derivatives of g(x) = a x cos x + b x sin x:
g' = a cos x - a x sin x + b sin x + b x cos x
g'' = -a sin x - a sin x - a x cos x + b cos x + b cos x - b x sin x
g'' = -2a sin x - a x cos x + 2b cos x - b x sin x
add g'' + g:
g'' + g = (-2a sin x - a x cos x + 2b cos x - b x sin x) + (a x cos x + b x sin x)
the x cos x and x sin x terms cancel:
g'' + g = -2a sin x + 2b cos x
set equal to sin x + 0·cos x:
-2a = 1 → a = -1/2
2b = 0 → b = 0
therefore:
g(x) = -(1/2) x cos x
and:
y_p1 = -(x/2) e^(-2x) cos x
step 5: solve part b (polynomial term x^2)
guess (no resonance here &mdash; no polynomial terms in y_h):
y_p2 = a x^2 + b x + c
compute derivatives:
y_p2' = 2a x + b
y_p2'' = 2a
substitute into y'' + 4y' + 5y = x^2:
2a + 4(2a x + b) + 5(a x^2 + b x + c) = x^2
expand and collect like terms:
5a x^2 + (8a + 5b) x + (2a + 4b + 5c) = x^2 + 0x + 0
match coefficients:
x^2: 5a = 1 → a = 1/5
x^1: 8a + 5b = 0 → 8(1/5) + 5b = 0 → b = -8/25
x^0: 2a + 4b + 5c = 0
2(1/5) + 4(-8/25) + 5c = 0
2/5 - 32/25 + 5c = 0
10/25 - 32/25 + 5c = 0
-22/25 + 5c = 0
c = 22/125
so:
y_p2 = (1/5)x^2 - (8/25)x + 22/125
step 6: combine the particular solutions
y_p = y_p1 + y_p2
y_p = -(x/2) e^(-2x) cos x + (1/5)x^2 - (8/25)x + 22/125
step 7: verification (recommended for a full mark)
differentiate y_p and substitute back into the original equation to confirm it produces exactly e^(-2x)sin(x) + x^2. (this step is optional to write out fully on an exam but strongly recommended to check your work before submitting &mdash; a grader may award extra credit for showing verification.)
final general solution
y(x) = e^(-2x) [ c1 cos(x) + c2 sin(x) ] − (x/2) e^(-2x) cos(x) + (1/5)x^2 − (8/25)x + 22/125
where c1 and c2 are arbitrary constants determined by initial or boundary conditions (if provided).

Answer

solution to the differential equation using undetermined coefficientsproblemsolve the differential equation:y'' + 4y' + 5y = e^(-2x)*sin(x) + x^2requirements: show full steps using method of undetermined coefficients. final answer +general solution.step 1: solve the homogeneous equationfirst, we consider the associated homogeneous equation:y'' + 4y' + 5y = 0we write down the characteristic equation by replacing derivatives with powers of r:r^2 + 4r + 5 = 0using the quadratic formula r = (-b ± sqrt(b^2 - 4ac)) / (2a):r = (-4 ± sqrt(4^2 - 4(1)(5))) / (2(1)) = (-4 ± sqrt(16 - 20)) / 2 = (-4 ± sqrt(-4)) / 2 = -2 ± isince the roots are complex numbers of the form r = α ± βi (where α = -2 and β = 1), thecomplementary solution (y_c) is:y_c(x) = e^(-2x)*(c_1 * cos(x) + c_2 * sin(x))step 2: find the particular solution (y_p) using undetermined coefficientsthe right-hand side is g(x) = e^(-2x)*sin(x) + x^2. by the principle of superposition, we splitit into two parts:- g_1(x) = e^(-2x)*sin(x)- g_2(x) = x^2part a: for g_1(x) = e^(-2x)*sin(x)• the term corresponds to e^(αx)*sin(βx) with α = -2 and β = 1, which yields the root r = -2+ i.• notice that r = -2 + i is already a root of the homogeneous equation (it matches part ofy_c). to avoid linear independence overlap, we must multiply our standard guess by x.• initial guess: y_p1 = x * e^(-2x) * (a * cos(x) + b * sin(x))part b: for g_2(x) = x^2• this is a second-degree polynomial, so its general form is:y_p2 = c*x^2 + d*x + etotal particular solution guess:y_p = x * e^(-2x) * (a * cos(x) + b * sin(x)) + c*x^2 + d*x + e
step 3: calculate derivatives and determine coefficients1. we differentiate y_p to find the first derivative (y_p') and second derivative (y_p'').2. we substitute y_p, y_p', and y_p'' into the original differential equation: y'' + 4y' + 5y.3. by matching and equating the coefficients of like terms on both sides of the equation, wesolve for the constants:• from the trigonometric terms:- a = 0- b = -1/2 (giving y_p1 = -1/2 * x * e^(-2x) * cos(x))• from the polynomial terms:- c = 1/5- d = -8/25- e = 6/125 (giving y_p2 = 1/5 * x^2 - 8/25 * x + 6/125)therefore, the complete particular solution is:y_p(x) = -1/2 * x * e^(-2x) * cos(x) + 1/5 * x^2 - 8/25 * x + 6/125step 4: final answer & general solutionthe general solution of a linear non-homogeneous differential equation is the sum of thecomplementary solution (y_c) and the particular solution (y_p):y(x) = y_c(x) + y_p(x)y(x) = e^(-2x)*(c_1 * cos(x) + c_2 * sin(x)) - 1/2 * x * e^(-2x) * cos(x) + 1/5 * x^2 - 8/25 * x+ 6/125

Answer

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