The solution consists of two parts:
f ( t ) = f h o m ( t ) + f p a r ( t ) f h o m ( t ) − solution of a homogeneous equation 2 f ′ ′ ( t ) − 3 f ′ ( t ) + f ( t ) = 0 f p a r ( t ) − particular solution of an inhomogeneous equation 2 f ′ ′ ( t ) − 3 f ′ ( t ) + f ( t ) = t ⋅ e t f(t)=f_{hom}(t)+f_{par}(t)\\[0.3cm]
f_{hom}(t)-\text{solution of a homogeneous equation}\\[0.3cm]
2f''(t)-3f'(t)+f(t)=0\\
f_{par}(t)-\text{particular solution of an inhomogeneous equation}\\[0.3cm]
2f''(t)-3f'(t)+f(t)=t\cdot e^t f ( t ) = f h o m ( t ) + f p a r ( t ) f h o m ( t ) − solution of a homogeneous equation 2 f ′′ ( t ) − 3 f ′ ( t ) + f ( t ) = 0 f p a r ( t ) − particular solution of an inhomogeneous equation 2 f ′′ ( t ) − 3 f ′ ( t ) + f ( t ) = t ⋅ e t
1 STEP: We solve the homogeneous equation
2 f ′ ′ ( t ) − 3 f ′ ( t ) + f ( t ) = 0 2f''(t)-3f'(t)+f(t)=0 2 f ′′ ( t ) − 3 f ′ ( t ) + f ( t ) = 0
We will seek a solution in the form f ( t ) = e k t f(t)=e^{kt} f ( t ) = e k t
f ( t ) = e k t ⟶ { f ′ ( t ) = k ⋅ e k t f ′ ′ ( t ) = k 2 ⋅ e k t ⟶ 2 ⋅ k 2 ⋅ e k t − 3 ⋅ k ⋅ e k t + e k t = 0 e k t ⋅ ( 2 k 2 − 3 k + 1 ) = 0 ⟶ 2 k 2 − 3 k + 1 = 0 ⏟ quadratic equation D = ( − 3 ) 2 − 4 ⋅ 2 ⋅ 1 = 1 = 1 [ k 1 = − ( − 3 ) + 1 2 ⋅ 2 = 1 k 2 = − ( − 3 ) − 1 2 ⋅ 2 = 1 2 f h o m ( t ) = C 1 ⋅ e t + C 2 ⋅ e t / 2 f(t)=e^{kt}\longrightarrow
\left\{\begin{array}{l}
f'(t)=k\cdot e^{kt}\\[0.3cm]
f''(t)=k^2\cdot e^{kt}
\end{array}\right.\longrightarrow\\[0.3cm]
2\cdot k^2\cdot e^{kt}-3\cdot k\cdot e^{kt}+e^{kt}=0\\[0.3cm]
e^{kt}\cdot\left(2k^2-3k+1\right)=0\longrightarrow\underbrace{2k^2-3k+1=0}_{\text{quadratic equation}}\\[0.3cm]
\sqrt{D}=\sqrt{\left(-3\right)^2-4\cdot 2\cdot1}=\sqrt{1}=1\\[0.3cm]
\left[\begin{array}{l}
k_1=\displaystyle\frac{-(-3)+1}{2\cdot2}=1\\[0.3cm]
k_2=\displaystyle\frac{-(-3)-1}{2\cdot2}=\displaystyle\frac{1}{2}
\end{array}\right.\\[0.3cm]
\boxed{f_{hom}(t)=C_1\cdot e^t+C_2\cdot e^{t/2}} f ( t ) = e k t ⟶ { f ′ ( t ) = k ⋅ e k t f ′′ ( t ) = k 2 ⋅ e k t ⟶ 2 ⋅ k 2 ⋅ e k t − 3 ⋅ k ⋅ e k t + e k t = 0 e k t ⋅ ( 2 k 2 − 3 k + 1 ) = 0 ⟶ quadratic equation 2 k 2 − 3 k + 1 = 0 D = ( − 3 ) 2 − 4 ⋅ 2 ⋅ 1 = 1 = 1 ⎣ ⎡ k 1 = 2 ⋅ 2 − ( − 3 ) + 1 = 1 k 2 = 2 ⋅ 2 − ( − 3 ) − 1 = 2 1 f h o m ( t ) = C 1 ⋅ e t + C 2 ⋅ e t /2
2 STEP: We solve the inhomogeneous equation
2 f ′ ′ ( t ) − 3 f ′ ( t ) + f ( t ) = t ⋅ e t 2f''(t)-3f'(t)+f(t)=t\cdot e^t 2 f ′′ ( t ) − 3 f ′ ( t ) + f ( t ) = t ⋅ e t
Minimum required theory: when solving inhomogeneous equations of the form
F ( f , f ′ , f ′ ′ , f ′ ′ ′ , … ) = g ( t ) F\left(f,f',f'',f''',\ldots\right)=g(t) F ( f , f ′ , f ′′ , f ′′′ , … ) = g ( t )
if the right side is g ( t ) = P m ( t ) ⋅ e α t , where P m ( t ) − polynomial g(t)=P_m(t)\cdot e^{\alpha t},\,\,\text{where}\,\,\, P_m(t)-\text{polynomial} g ( t ) = P m ( t ) ⋅ e α t , where P m ( t ) − polynomial
[ P m ~ ( t ) ⋅ e α t − α not the root of the characteristic equation x s P m ~ ( t ) ⋅ e α t − α the root of the characteristic equation \left[\begin{array}{l}
\widetilde{P_m}(t)\cdot e^{\alpha t}-\alpha\,\,\text{not the root of the characteristic equation}\\[0.3cm]
x^s\widetilde{P_m}(t)\cdot e^{\alpha t}-\alpha\,\,\text{the root of the characteristic equation}
\end{array}\right. ⎣ ⎡ P m ( t ) ⋅ e α t − α not the root of the characteristic equation x s P m ( t ) ⋅ e α t − α the root of the characteristic equation
In our case,
g ( t ) = t ⋅ e 1 ⋅ t ⟶ k 1 = 1 − multiplicity root 1 f p a r ( t ) = t ⋅ ( A t + B ) ⋅ e t ≡ ( A t 2 + B t ) ⋅ e t ⟶ { f ′ ( t ) = ( A t 2 + B t ) ⋅ e t + ( 2 A t + B ) ⋅ e t f ′ ( t ) = ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t f ′ ′ ( t ) = ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t + ( 2 A t + ( 2 A + B ) ) ⋅ e t f ′ ′ ( t ) = ( A t 2 + ( 4 A + B ) t + ( 2 A + 2 B ) ) ⋅ e t g(t)=t\cdot e^{1\cdot t}\longrightarrow k_1=1-\text{multiplicity root 1}\\[0.3cm]
f_{par}(t)=t\cdot\left(At+B\right)\cdot e^t\equiv\left(At^2+Bt\right)\cdot e^t\longrightarrow\\[0.3cm]
\left\{\begin{array}{l}
f'(t)=\left(At^2+Bt\right)\cdot e^t+\left(2At+B\right)\cdot e^t\\[0.3cm]
f'(t)=\left(At^2+(2A+B)t+B\right)\cdot e^t\\[0.3cm]
f''(t)=\left(At^2+(2A+B)t+B\right)\cdot e^t+\left(2At+(2A+B)\right)\cdot e^t\\[0.3cm]
f''(t)=\left(At^2+(4A+B)t+(2A+2B)\right)\cdot e^t
\end{array}\right. g ( t ) = t ⋅ e 1 ⋅ t ⟶ k 1 = 1 − multiplicity root 1 f p a r ( t ) = t ⋅ ( A t + B ) ⋅ e t ≡ ( A t 2 + Bt ) ⋅ e t ⟶ ⎩ ⎨ ⎧ f ′ ( t ) = ( A t 2 + Bt ) ⋅ e t + ( 2 A t + B ) ⋅ e t f ′ ( t ) = ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t f ′′ ( t ) = ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t + ( 2 A t + ( 2 A + B ) ) ⋅ e t f ′′ ( t ) = ( A t 2 + ( 4 A + B ) t + ( 2 A + 2 B ) ) ⋅ e t
Substitute in the original equation
2 ( A t 2 + ( 4 A + B ) t + ( 2 A + 2 B ) ) ⋅ e t − − 3 ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t + ( A t 2 + B t ) ⋅ e t = t ⋅ e t ( 2 A − 3 A + A ) t 2 + ( 2 ( 4 A + B ) − 3 ( 2 A + B ) + B ) t + + ( 2 ( 2 A + 2 B ) − 3 B ) = t 0 ⋅ t 2 + ( 2 A ) ⋅ t + ( 4 A + B ) = t ⟶ { 2 A = 1 4 A + B = 0 ⟶ { A = 1 2 B = − 4 A = − 2 2\left(At^2+(4A+B)t+(2A+2B)\right)\cdot e^t-\\[0.3cm]
-3\left(At^2+(2A+B)t+B\right)\cdot e^t+\left(At^2+Bt\right)\cdot e^t=t\cdot e^t\\[0.3cm]
(2A-3A+A)t^2+(2(4A+B)-3(2A+B)+B)t+\\[0.3cm]
+(2(2A+2B)-3B)=t\\[0.3cm]
0\cdot t^2+(2A)\cdot t+(4A+B)=t\longrightarrow\\[0.3cm]
\left\{\begin{array}{l}
2A=1\\[0.3cm]
4A+B=0
\end{array}\right.
\longrightarrow
\left\{\begin{array}{l}
A=\displaystyle\frac{1}{2}\\[0.3cm]
B=-4A=-2
\end{array}\right. 2 ( A t 2 + ( 4 A + B ) t + ( 2 A + 2 B ) ) ⋅ e t − − 3 ( A t 2 + ( 2 A + B ) t + B ) ⋅ e t + ( A t 2 + Bt ) ⋅ e t = t ⋅ e t ( 2 A − 3 A + A ) t 2 + ( 2 ( 4 A + B ) − 3 ( 2 A + B ) + B ) t + + ( 2 ( 2 A + 2 B ) − 3 B ) = t 0 ⋅ t 2 + ( 2 A ) ⋅ t + ( 4 A + B ) = t ⟶ { 2 A = 1 4 A + B = 0 ⟶ ⎩ ⎨ ⎧ A = 2 1 B = − 4 A = − 2
Conclusion,
f p a r ( t ) = ( t 2 2 − 2 t ) ⋅ e t \boxed{f_{par}(t)=\left(\frac{t^2}{2}-2t\right)\cdot e^t} f p a r ( t ) = ( 2 t 2 − 2 t ) ⋅ e t
General solution,
f ( t ) = C 1 ⋅ e t + C 2 ⋅ e t / 2 + ( t 2 2 − 2 t ) ⋅ e t \boxed{f(t)=C_1\cdot e^t+C_2\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t} f ( t ) = C 1 ⋅ e t + C 2 ⋅ e t /2 + ( 2 t 2 − 2 t ) ⋅ e t
The constants are determined from the initial conditions
f ( 0 ) = 1 = C 1 ⋅ e 0 + C 2 ⋅ e 0 / 2 + ( 0 2 2 − 2 ⋅ 0 ) ⋅ e 0 C 1 + C 2 = 1 f ′ ( t ) = C 1 ⋅ e t + C 2 2 ⋅ e t / 2 + ( 2 t − 2 ) ⋅ e t + ( t 2 2 − 2 t ) ⋅ e t f ′ ( 0 ) = 3 2 = C 1 ⋅ e 0 + C 2 2 ⋅ e 0 / 2 + ( 2 ⋅ 0 − 2 ) ⋅ e 0 + ( 0 2 2 − 2 ⋅ 0 ) ⋅ e 0 C 1 + C 2 2 − 2 = 3 2 f(0)=1=C_1\cdot e^0+C_2\cdot e^{0/2}+\left(\frac{0^2}{2}-2\cdot0\right)\cdot e^0\\[0.3cm]
\boxed{C_1+C_2=1}\\[0.3cm]
f'(t)=C_1\cdot e^t+\frac{C_2}{2}\cdot e^{t/2}+(2t-2)\cdot e^t+\left(\frac{t^2}{2}-2t\right)\cdot e^t\\[0.3cm]
f'(0)=\frac{3}{2}=C_1\cdot e^0+\frac{C_2}{2}\cdot e^{0/2}+(2\cdot0-2)\cdot e^0+\left(\frac{0^2}{2}-2\cdot0\right)\cdot e^0\\[0.3cm]
\boxed{C_1+\frac{C_2}{2}-2=\frac{3}{2}} f ( 0 ) = 1 = C 1 ⋅ e 0 + C 2 ⋅ e 0/2 + ( 2 0 2 − 2 ⋅ 0 ) ⋅ e 0 C 1 + C 2 = 1 f ′ ( t ) = C 1 ⋅ e t + 2 C 2 ⋅ e t /2 + ( 2 t − 2 ) ⋅ e t + ( 2 t 2 − 2 t ) ⋅ e t f ′ ( 0 ) = 2 3 = C 1 ⋅ e 0 + 2 C 2 ⋅ e 0/2 + ( 2 ⋅ 0 − 2 ) ⋅ e 0 + ( 2 0 2 − 2 ⋅ 0 ) ⋅ e 0 C 1 + 2 C 2 − 2 = 2 3
It remains to solve the system of equations
{ C 1 + C 2 = 1 C 1 + C 2 2 − 2 = 3 2 ∣ × ( 2 ) ⟶ { C 2 = 1 − C 1 2 C 1 + C 2 − 4 = 3 ⟶ { C 2 = 1 − C 1 2 C 1 + 1 − C 1 − 4 = 3 ⟶ { C 2 = − 5 C 1 = 6 \left\{\begin{array}{l}
C_1+C_2=1\\[0.3cm]
\left.C_1+\displaystyle\frac{C_2}{2}-2=\displaystyle\frac{3}{2}\right|\times(2)
\end{array}\right.\longrightarrow\left\{\begin{array}{l}
C_2=1-C_1\\[0.3cm]
2C_1+C_2-4=3\end{array}\right.\longrightarrow\\[0.3cm]
\left\{\begin{array}{l}
C_2=1-C_1\\[0.3cm]
2C_1+1-C_1-4=3\end{array}\right.\longrightarrow
\left\{\begin{array}{l}
C_2=-5\\[0.3cm]
C_1=6\end{array}\right. ⎩ ⎨ ⎧ C 1 + C 2 = 1 C 1 + 2 C 2 − 2 = 2 3 ∣ ∣ × ( 2 ) ⟶ { C 2 = 1 − C 1 2 C 1 + C 2 − 4 = 3 ⟶ { C 2 = 1 − C 1 2 C 1 + 1 − C 1 − 4 = 3 ⟶ { C 2 = − 5 C 1 = 6
Conclusion, solution of IVT is
f ( t ) = 6 ⋅ e t − 5 ⋅ e t / 2 + ( t 2 2 − 2 t ) ⋅ e t \boxed{f(t)=6\cdot e^t-5\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t} f ( t ) = 6 ⋅ e t − 5 ⋅ e t /2 + ( 2 t 2 − 2 t ) ⋅ e t
ANSWER
f ( t ) = 6 ⋅ e t − 5 ⋅ e t / 2 + ( t 2 2 − 2 t ) ⋅ e t f(t)=6\cdot e^t-5\cdot e^{t/2}+\left(\frac{t^2}{2}-2t\right)\cdot e^t f ( t ) = 6 ⋅ e t − 5 ⋅ e t /2 + ( 2 t 2 − 2 t ) ⋅ e t
#339914 on Dec 2023
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