Consider the differential equation,
y ′ ′ − y ′ = 0 L e t y = ∑ n = 0 ∞ c n x n , y^{\prime \prime}-y^{\prime}=0\ Let\ y=\sum_{n=0}^{\infty} c_{n} x^{n}, y ′′ − y ′ = 0 L e t y = ∑ n = 0 ∞ c n x n ,
t h e n y ′ = ∑ n = 1 ∞ n c n x n − 1 a n d y ′ ′ = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 then y^{\prime}=\sum_{n=1}^{\infty} n c_{n} x^{n-1} and y^{\prime \prime}=\sum_{n=2}^{\infty} n(n-1) c_{n} x^{n-2} t h e n y ′ = ∑ n = 1 ∞ n c n x n − 1 an d y ′′ = ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2
Substituting in the differential equation we get
∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 − ∑ n = 1 ∞ n c n x n − 1 = 0 ⇒ ∑ n = 2 ∞ n ( n − 1 ) c n x n − 2 ⏟ k = n − 2 − ∑ n = 1 ∞ n c n x n − 1 ⏟ k = n − 1 = 0 ∑ k = 0 ∞ ( k + 2 ) ( k + 1 ) c k + 2 x k − ∑ k = 0 ∞ ( k + 1 ) c k + 1 x k = 0 ∑ k = 0 ∞ [ ( k + 2 ) ( k + 1 ) c k + 2 − ( k + 1 ) c k + 1 ] x k = 0 \begin{aligned}
\sum_{n=2}^{\infty} & n(n-1) c_{n} x^{n-2}-\sum_{n=1}^{\infty} n c_{n} x^{n-1}=0 \\
\Rightarrow & \sum_{n=2}^{\infty} \underbrace{n(n-1) c_{n} x^{n-2}}_{k=n-2}-\sum_{n=1}^{\infty} \underbrace{n c_{n} x^{n-1}}_{k=n-1}=0 \\
\sum_{k=0}^{\infty}(k+2)(k+1) c_{k+2} x^{k}-\sum_{k=0}^{\infty}(k+1) c_{k+1} x^{k}=0 \\
& \sum_{k=0}^{\infty}\left[(k+2)(k+1) c_{k+2}-(k+1) c_{k+1}\right] x^{k}=0
\end{aligned} n = 2 ∑ ∞ ⇒ k = 0 ∑ ∞ ( k + 2 ) ( k + 1 ) c k + 2 x k − k = 0 ∑ ∞ ( k + 1 ) c k + 1 x k = 0 n ( n − 1 ) c n x n − 2 − n = 1 ∑ ∞ n c n x n − 1 = 0 n = 2 ∑ ∞ k = n − 2 n ( n − 1 ) c n x n − 2 − n = 1 ∑ ∞ k = n − 1 n c n x n − 1 = 0 k = 0 ∑ ∞ [ ( k + 2 ) ( k + 1 ) c k + 2 − ( k + 1 ) c k + 1 ] x k = 0
Compare the coefficients on each side to get,
( k + 2 ) ( k + 1 ) c k + 2 − ( k + 1 ) c k + 1 = 0 , k = 0 , 1 , 2 , … c k + 2 = 1 k + 2 c k + 1 , k = 0 , 1 , 2 , … \begin{aligned}
(k+2)(k+1) c_{k+2}-(k+1) c_{k+1} &=0, k=0,1,2, \ldots \\
c_{k+2} &=\frac{1}{k+2} c_{k+1}, k=0,1,2, \ldots
\end{aligned} ( k + 2 ) ( k + 1 ) c k + 2 − ( k + 1 ) c k + 1 c k + 2 = 0 , k = 0 , 1 , 2 , … = k + 2 1 c k + 1 , k = 0 , 1 , 2 , …
From the relation c k + 2 = 1 k + 2 c k + 1 , k = 0 , 1 , 2 , … c_{k+2}=\frac{1}{k+2} c_{k+1}, k=0,1,2, \ldots c k + 2 = k + 2 1 c k + 1 , k = 0 , 1 , 2 , …
c 2 = 1 0 + 2 c 0 + 1 = 1 2 c 1 c 3 = 1 1 + 2 c 1 + 1 = 1 3 ( 1 2 c 1 ) = 1 6 c 1 c 4 = 1 2 + 2 c 2 + 1 = 1 4 ( 1 6 c 1 ) = 1 24 c 1 \begin{aligned}
c_{2} &=\frac{1}{0+2} c_{0+1} \\
&=\frac{1}{2} c_{1} \\
c_{3} &=\frac{1}{1+2} c_{1+1} \\
&=\frac{1}{3}\left(\frac{1}{2} c_{1}\right) \\
&=\frac{1}{6} c_{1} \\
c_{4} &=\frac{1}{2+2} c_{2+1} \\
&=\frac{1}{4}\left(\frac{1}{6} c_{1}\right) \\
&=\frac{1}{24} c_{1}
\end{aligned} c 2 c 3 c 4 = 0 + 2 1 c 0 + 1 = 2 1 c 1 = 1 + 2 1 c 1 + 1 = 3 1 ( 2 1 c 1 ) = 6 1 c 1 = 2 + 2 1 c 2 + 1 = 4 1 ( 6 1 c 1 ) = 24 1 c 1
Rewrite the solution y 2 y_{2} y 2
y 2 = x + 1 2 ! x 2 + 1 3 ! x 3 + 1 4 ! x 4 + ⋯ = − 1 + 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + 1 4 ! x 4 + ⋯ = − 1 + ( 1 + x + 1 2 ! x 2 + 1 3 ! x 3 + 1 4 ! x 4 + ⋯ ) = − 1 + ∑ n = 0 ∞ x n n ! = − 1 + e x \begin{aligned}
y_{2} &=x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots \\
&=-1+1+x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots \\
&=-1+\left(1+x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\frac{1}{4 !} x^{4}+\cdots\right) \\
&=-1+\sum_{n=0}^{\infty} \frac{x^{n}}{n !} \\
&=-1+e^{x}
\end{aligned} y 2 = x + 2 ! 1 x 2 + 3 ! 1 x 3 + 4 ! 1 x 4 + ⋯ = − 1 + 1 + x + 2 ! 1 x 2 + 3 ! 1 x 3 + 4 ! 1 x 4 + ⋯ = − 1 + ( 1 + x + 2 ! 1 x 2 + 3 ! 1 x 3 + 4 ! 1 x 4 + ⋯ ) = − 1 + n = 0 ∑ ∞ n ! x n = − 1 + e x
Therefore, the two solutions are y 1 = 1 , y 2 = − 1 + e x y_{1}=1, y_{2}=-1+e^{x} y 1 = 1 , y 2 = − 1 + e x
Finally general solution to the given differential equation is,
y ( x ) = c 0 ⋅ 1 + c 1 ⋅ ( e x − 1 ) y(x)=c_{0} \cdot 1+c_{1} \cdot\left(e^{x}-1\right) y ( x ) = c 0 ⋅ 1 + c 1 ⋅ ( e x − 1 )
#339914 on Dec 2023