Equation #1
y ′ ′ ′ + y ′ ′ = y + x 5 y'''+y''=y+x^5 y ′′′ + y ′′ = y + x 5 1 STEP: Solve a homogeneous equation
y ′ ′ ′ + y ′ ′ − y = 0 y'''+y''-y=0\\ y ′′′ + y ′′ − y = 0 We are looking for a solution in the form
y ( x ) = e k x ⟶ y ′ ′ = k 2 ⋅ e k x a n d y ′ ′ ′ = k 3 ⋅ e k x y(x)=e^{kx}\longrightarrow\,y''=k^2\cdot e^{kx}\,\,\,and\,\,\,y'''=k^3\cdot e^{kx} y ( x ) = e k x ⟶ y ′′ = k 2 ⋅ e k x an d y ′′′ = k 3 ⋅ e k x Then,
k 3 ⋅ e k x + k 2 ⋅ e k x − e k x = 0 ⟶ k 3 + k 2 − 1 = 0 k^3\cdot e^{kx}+k^2\cdot e^{kx}-e^{kx}=0\longrightarrow\\
k^3+k^2-1=0 k 3 ⋅ e k x + k 2 ⋅ e k x − e k x = 0 ⟶ k 3 + k 2 − 1 = 0 the cubic equation has three solutions: one real and two complex
k 1 = 1 3 ( − 1 + 25 2 − 3 69 2 3 + 25 + 3 69 2 3 ) k 2 = − 1 3 − 1 − i 3 6 ( 25 2 − 3 69 2 3 ) − 1 + i 3 6 ( 25 + 3 69 2 3 ) k 3 = − 1 3 − 1 + i 3 6 ( 25 2 − 3 69 2 3 ) − 1 − i 3 6 ( 25 + 3 69 2 3 ) k_1=\frac{1}{3}\left(-1+\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}+\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm]
k_2=-\frac{1}{3}-\frac{1-i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}\right)-\frac{1+
i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm]
k_3=-\frac{1}{3}-\frac{1+i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25}{2}-\frac{3\sqrt{69}}{2}}\right)-\frac{1-
i\sqrt{3}}{6}\left(\sqrt[3]{\frac{25+3\sqrt{69}}{2}}\right)\\[0.5cm] k 1 = 3 1 ⎝ ⎛ − 1 + 3 2 25 − 2 3 69 + 3 2 25 + 3 69 ⎠ ⎞ k 2 = − 3 1 − 6 1 − i 3 ⎝ ⎛ 3 2 25 − 2 3 69 ⎠ ⎞ − 6 1 + i 3 ⎝ ⎛ 3 2 25 + 3 69 ⎠ ⎞ k 3 = − 3 1 − 6 1 + i 3 ⎝ ⎛ 3 2 25 − 2 3 69 ⎠ ⎞ − 6 1 − i 3 ⎝ ⎛ 3 2 25 + 3 69 ⎠ ⎞ Then,
y ( x ) = C 1 ⋅ e k 1 x + C 2 ⋅ e k 2 x + C 3 ⋅ e k 3 x − h o m o g e n e o u s s o l u t i o n \boxed{y(x)=C_1\cdot e^{k_1x}+C_2\cdot e^{k_2x}+C_3\cdot e^{k_3x}-homogeneous\,\,\,solution} y ( x ) = C 1 ⋅ e k 1 x + C 2 ⋅ e k 2 x + C 3 ⋅ e k 3 x − h o m o g e n eo u s so l u t i o n 2 STEP: Solve an Inhomogeneous Equation
We are looking for a solution in the form
y ( x ) = A x 5 + B x 4 + C x 3 + D x 2 + E x + F y(x)=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F y ( x ) = A x 5 + B x 4 + C x 3 + D x 2 + E x + F Then,
y ′ ′ = 20 A x 3 + 12 B x 2 + 6 C x + 2 D y ′ ′ ′ = 60 A x 2 + 24 B x + 6 C y''=20Ax^3+12Bx^2+6Cx+2D\\[0.3cm]
y'''=60Ax^2+24Bx+6C y ′′ = 20 A x 3 + 12 B x 2 + 6 C x + 2 D y ′′′ = 60 A x 2 + 24 B x + 6 C Substitute in the original equation
60 A x 2 + 24 B x + 6 C + 20 A x 3 + 12 B x 2 + 6 C x + 2 D = = A x 5 + B x 4 + C x 3 + D x 2 + E x + F + x 5 ⟶ ( A + 1 ) x 5 + B x 4 + ( C − 20 A ) x 3 + + ( D − 60 A − 12 B ) x 2 + ( E − 24 B − 6 C ) x + ( F − 6 C − 2 D ) = 0 A + 1 = 0 ⟶ A = − 1 B = 0 C − 20 A = 0 ⟶ C = − 20 D − 60 A − 12 B = 0 ⟶ D = − 60 E − 24 B − 6 C = 0 ⟶ E = − 120 F − − 6 C − 2 D = 0 ⟶ F = − 240 60Ax^2+24Bx+6C+20Ax^3+12Bx^2+6Cx+2D=\\[0.3cm]
=Ax^5+Bx^4+Cx^3+Dx^2+Ex+F+x^5\longrightarrow\\[0.3cm]
(A+1)x^5+Bx^4+(C-20A)x^3+\\[0.3cm]
+(D-60A-12B)x^2+(E-24B-6C)x+(F-6C-2D)=0\\[0.3cm]
A+1=0\longrightarrow\boxed{A=-1}\\[0.3cm]
\boxed{B=0}\\[0.3cm]
C-20A=0\longrightarrow\boxed{C=-20}\\[0.3cm]
D-60A-12B=0\longrightarrow\boxed{D=-60}\\[0.3cm]
E-24B-6C=0\longrightarrow\boxed{E=-120}\\[0.3cm]
F--6C-2D=0\longrightarrow\boxed{F=-240}\\[0.3cm] 60 A x 2 + 24 B x + 6 C + 20 A x 3 + 12 B x 2 + 6 C x + 2 D = = A x 5 + B x 4 + C x 3 + D x 2 + E x + F + x 5 ⟶ ( A + 1 ) x 5 + B x 4 + ( C − 20 A ) x 3 + + ( D − 60 A − 12 B ) x 2 + ( E − 24 B − 6 C ) x + ( F − 6 C − 2 D ) = 0 A + 1 = 0 ⟶ A = − 1 B = 0 C − 20 A = 0 ⟶ C = − 20 D − 60 A − 12 B = 0 ⟶ D = − 60 E − 24 B − 6 C = 0 ⟶ E = − 120 F − − 6 C − 2 D = 0 ⟶ F = − 240 Conclusion,
y ( x ) = − x 5 − 20 x 3 − 60 x 2 − 120 x − 240 − i n h o m o g e n e o u s s o l u t i o n y(x)=-x^5-20x^3-60x^2-120x-240-inhomogeneous\,\,\,solution y ( x ) = − x 5 − 20 x 3 − 60 x 2 − 120 x − 240 − inh o m o g e n eo u s so l u t i o n
General conclusion:
y ( x ) = C 1 e k 1 x + C 2 e k 2 x + C 3 e k 3 x − x 5 − 20 x 3 − 60 x 2 − 120 x − 240 \boxed{y(x)=C_1e^{k_1x}+C_2e^{k_2x}+C_3e^{k_3x}-x^5-20x^3-60x^2-120x-240} y ( x ) = C 1 e k 1 x + C 2 e k 2 x + C 3 e k 3 x − x 5 − 20 x 3 − 60 x 2 − 120 x − 240
ANSWER
y ( x ) = C 1 e k 1 x + C 2 e k 2 x + C 3 e k 3 x − x 5 − 20 x 3 − 60 x 2 − 120 x − 240 y(x)=C_1e^{k_1x}+C_2e^{k_2x}+C_3e^{k_3x}-x^5-20x^3-60x^2-120x-240 y ( x ) = C 1 e k 1 x + C 2 e k 2 x + C 3 e k 3 x − x 5 − 20 x 3 − 60 x 2 − 120 x − 240
Equation #2:
y ⋅ y ′ ′ = y ′ + 8 x 3 y\cdot y''=y'+8x^3 y ⋅ y ′′ = y ′ + 8 x 3 1 STEP: Solve a homogeneous equation
y ⋅ y ′ ′ = y ′ ⟶ y ′ ′ = y ′ y ⟶ d d x ( y ′ ) = d d x ( ln ( y ) ) ⟶ y ′ = ln ( C y ) ⟶ d y d x = ln ( C y ) ⟶ ∫ d y ln ( C y ) = ∫ 1 d x y\cdot y''=y'\longrightarrow y''=\frac{y'}{y}\longrightarrow\\[0.3cm]
\frac{d}{dx}\left(y'\right)=\frac{d}{dx}\left(\ln(y)\right)\longrightarrow y'=\ln(Cy)\longrightarrow\\[0.3cm]
\frac{dy}{dx}=\ln(Cy)\longrightarrow\int\frac{dy}{\ln(Cy)}=\int1dx y ⋅ y ′′ = y ′ ⟶ y ′′ = y y ′ ⟶ d x d ( y ′ ) = d x d ( ln ( y ) ) ⟶ y ′ = ln ( C y ) ⟶ d x d y = ln ( C y ) ⟶ ∫ ln ( C y ) d y = ∫ 1 d x Since
∫ d y ln ( C y ) i s t h e i n t e g r a l l o g a r i t h m , n o t a n e l e m e n t a r y f u n c t i o n \int\frac{dy}{\ln(Cy)} \,is\, the\,integral\,logarithm, not\,an\,elementary\,function ∫ ln ( C y ) d y i s t h e in t e g r a l l o g a r i t hm , n o t an e l e m e n t a ry f u n c t i o n
This means that we can’t write down the general solution of this equation in the usual elementary functions. The only thing we CAN write is
l i ( y ) = x + C \boxed{li(y)=x+C} l i ( y ) = x + C More information: https://en.wikipedia.org/wiki/Logarithmic_integral_function
#339914 on Dec 2023
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