y ′ ′ − 16 y ′ + 64 y = − 2 x 2 − 5 e 5 x y''-16y'+64y=-2x^2-5e^{5x} y ′′ − 16 y ′ + 64 y = − 2 x 2 − 5 e 5 x
Let us solve a characteristic equation:
k 2 − 16 k + 64 = 0 k^2-16k+64=0 k 2 − 16 k + 64 = 0
( k − 8 ) 2 = 0 (k-8)^2=0 ( k − 8 ) 2 = 0
k 1 = k 2 = 8 k_1= k_2=8 k 1 = k 2 = 8
The general solution is of the following form:
y ( x ) = e 8 x ( C 1 + C 2 x ) + y p ( x ) , y(x)=e^{8x}(C_1+C_2 x)+y_p(x), y ( x ) = e 8 x ( C 1 + C 2 x ) + y p ( x ) , y p ( x ) y_p(x) y p ( x ) C 1 , C 2 C_1, C_2 C 1 , C 2
The particular solution has the following form:
y p ( x ) = a x 2 + b x + c + d e 5 x y_p(x)=ax^2+bx+c+de^{5x} y p ( x ) = a x 2 + b x + c + d e 5 x a , b , c a,b,c a , b , c d d d
y p ′ = 2 a x + b + 5 d e 5 x y_p'=2ax+b+5de^{5x} y p ′ = 2 a x + b + 5 d e 5 x
y p ′ ′ = 2 a + 25 d e 5 x y_p''=2a+25de^{5x} y p ′′ = 2 a + 25 d e 5 x
Put founded derivatives in the differential equation:
2 a + 25 d e 5 x − 16 ( 2 a x + b + 5 d e 5 x ) + 64 ( a x 2 + b x + c + d e 5 x ) = − 2 x 2 − 5 e 5 x 2a+25de^{5x}-16(2ax+b+5de^{5x})+64(ax^2+bx+c+de^{5x})=-2x^2-5e^{5x} 2 a + 25 d e 5 x − 16 ( 2 a x + b + 5 d e 5 x ) + 64 ( a x 2 + b x + c + d e 5 x ) = − 2 x 2 − 5 e 5 x
2 a + 25 d e 5 x − 32 a x − 16 b − 80 d e 5 x + 64 a x 2 + 64 b x + 64 c + 64 d e 5 x = − 2 x 2 − 5 e 5 x 2a+25de^{5x}-32ax-16b-80de^{5x}+64ax^2+64bx+64c+64de^{5x}=-2x^2-5e^{5x} 2 a + 25 d e 5 x − 32 a x − 16 b − 80 d e 5 x + 64 a x 2 + 64 b x + 64 c + 64 d e 5 x = − 2 x 2 − 5 e 5 x
64 a x 2 + ( − 32 a + 64 b ) x + ( 2 a − 16 b + 64 c ) + 9 d e 5 x = − 2 x 2 − 5 e 5 x 64ax^2+(-32a+64b)x+(2a-16b+64c)+9de^{5x}=-2x^2-5e^{5x} 64 a x 2 + ( − 32 a + 64 b ) x + ( 2 a − 16 b + 64 c ) + 9 d e 5 x = − 2 x 2 − 5 e 5 x
Equating the corresponding coefficients, we obtain a system of equations:
{ 64 a = − 2 − 32 a + 64 b = 0 2 a − 16 b + 64 c = 0 9 d = − 5 \begin{cases} 64a=-2\\ -32a+64b=0\\2a-16b+64c=0\\9d=-5 \end{cases} ⎩ ⎨ ⎧ 64 a = − 2 − 32 a + 64 b = 0 2 a − 16 b + 64 c = 0 9 d = − 5
{ a = − 1 32 a = 2 b 64 c = − 2 a + 16 b 9 d = − 5 \begin{cases} a=-\frac{1}{32}\\ a=2b\\64c=-2a+16b\\9d=-5 \end{cases} ⎩ ⎨ ⎧ a = − 32 1 a = 2 b 64 c = − 2 a + 16 b 9 d = − 5
{ a = − 1 32 b = − 1 64 64 c = 1 16 − 1 4 d = − 5 9 \begin{cases} a=-\frac{1}{32}\\ b=-\frac{1}{64}\\64c=\frac{1}{16}-\frac{1}{4}\\d=-\frac{5}{9} \end{cases} ⎩ ⎨ ⎧ a = − 32 1 b = − 64 1 64 c = 16 1 − 4 1 d = − 9 5
{ a = − 1 32 b = − 1 64 c = − 3 1024 d = − 5 9 \begin{cases} a=-\frac{1}{32}\\ b=-\frac{1}{64}\\c=-\frac{3}{1024}\\d=-\frac{5}{9} \end{cases} ⎩ ⎨ ⎧ a = − 32 1 b = − 64 1 c = − 1024 3 d = − 9 5
Finally, the solution of the differential equation is the following:
y ( x ) = e 8 x ( C 1 + C 2 x ) − 1 32 x 2 − 1 64 x − 3 1024 − 5 9 e 5 x y(x)=e^{8x}(C_1+C_2 x)-\frac{1}{32}x^2-\frac{1}{64}x-\frac{3}{1024}-\frac{5}{9}e^{5x} y ( x ) = e 8 x ( C 1 + C 2 x ) − 32 1 x 2 − 64 1 x − 1024 3 − 9 5 e 5 x
#339914 on Dec 2023
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