hcju/proofwikiqa huggingface dataset:
Let $S \subseteq \N_{>0}$ denote the set of (strictly positive) natural numbers for which $(1)$ holds. === Basis for the Induction === We have: {{begin-eqn}} {{eqn | l = 2^1 - 1       | r = 2 - 1       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = 2^0       | c =  }} {{eqn | r = \sum_{j \mathop = 0}^{1 - 1} 2^j       | c =  }} {{end-eqn}} So $1 \in S$. This is our basis for the induction. === Induction Hypothesis === We now show that, if $k \in S$ is true, where $k \ge 1$, then it logically follows that $k + 1 \in S$. So this is our induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^{k - 1} 2^j = 2^k - 1$ Then we need to show: :$\displaystyle \sum_{j \mathop = 0}^k 2^j = 2^{k + 1} - 1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^k 2^j       | r = \sum_{j \mathop = 0}^{k - 1} 2^j + 2^k       | c =  }} {{eqn | r = 2^k - 1 + 2^k       | c = Induction Hypothesis }} {{eqn | r = 2 \times 2^k - 1       | c =  }} {{eqn | r = 2^{k + 1} - 1       | c =  }} {{end-eqn}} So $k \in S \implies k + 1 \in S$. It follows by the Principle of Finite Induction that $S = \N_{>0}$. That is: :$\displaystyle \forall n \in \N_{> 0}: \sum_{j \mathop = 0}^{n - 1} 2^j = 2^n - 1$ {{qed}}
The proof proceeds by complete induction. For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition: :$L_n < \paren {\dfrac 7 4}^n$ $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = L_1       | r = 1       | c =  }} {{eqn | o = <       | r = \dfrac 7 4       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. === Basis for the Induction === $\map P 2$ is the case: {{begin-eqn}} {{eqn | l = L_2       | r = 3       | c =  }} {{eqn | r = \dfrac {48} {16}       | c =  }} {{eqn | o = <       | r = \dfrac {49} {16}       | c =  }} {{eqn | r = \paren {\dfrac 7 4}^2       | c =  }} {{end-eqn}} Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P j$ is true, for all $j$ such that $0 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true. This is the induction hypothesis: :$L_k < \paren {\dfrac 7 4}^k$ from which it is to be shown that: :$L_{k + 1} < \paren {\dfrac 7 4}^{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = L_{k + 1}       | r = L_k + L_{k - 1}       | c =  }} {{eqn | o = <       | r = \paren {\dfrac 7 4}^k + \paren {\dfrac 7 4}^{k - 1}       | c = Induction Hypothesis }} {{eqn | r = \paren {\dfrac 7 4}^{k - 1} \paren {1 + \dfrac 7 4}       | c =  }} {{eqn | r = \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac {11} 4}       | c =  }} {{eqn | o = <       | r = \paren {\dfrac 7 4}^{k - 1} \paren {\dfrac 7 4}^2       | c =  }} {{eqn | r = \paren {\dfrac 7 4}^{k + 1}       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Second Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 1}: L_n < \paren {\dfrac 7 4}^n$ {{qed}}
{{begin-eqn}} {{eqn | l = n^3       | r = \sum_{k \mathop = 1}^n k^3 - \sum_{k \mathop = 1}^{n - 1} k^3       | c =  }} {{eqn | r = \paren {\frac {n^2 \paren {n + 1}^2} 4} - \paren {\frac {\paren {n - 1}^2 n^2} 4}       | c = Sum of Sequence of Cubes }} {{eqn | r = \frac {n^2 \paren {\paren {n + 1}^2 - \paren {n - 1}^2} } 4       | c =  }} {{eqn | r = \paren {\frac {n \paren {n + 1} } 2}^2 - \paren {\frac {n \paren {n - 1} } 2}^2       | c =  }} {{end-eqn}} {{qed}}
We note that: {{begin-eqn}} {{eqn | l = 1!       | r = 1       | c =  }} {{eqn | r = 1^2       | c =  }} {{eqn | l = 2!       | r = 2       | c =  }} {{eqn | o = <       | r = 4       | c =  }} {{eqn | r = 2^2       | c =  }} {{eqn | l = 3!       | r = 6       | c =  }} {{eqn | o = <       | r = 9       | c =  }} {{eqn | r = 3^2       | c =  }} {{end-eqn}} The proof then proceeds by induction. For all $n \in \Z_{\ge 4}$, let $\map P n$ be the proposition: :$n! > n^2$ === Basis for the Induction === $\map P 4$ is the case: {{begin-eqn}} {{eqn | l = 4!       | r = 24       | c =  }} {{eqn | o = >       | r = 16       | c =  }} {{eqn | r = 4^2       | c =  }} {{end-eqn}} Thus $\map P 4$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 4$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k! > k^2$ from which it is to be shown that: :$\paren {k + 1}! > \paren {k + 1}^2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \paren {k + 1}!       | r = \paren {k + 1} \times k!        | c =  }} {{eqn | o = >       | r = \paren {k + 1} \times k^2        | c = Induction Hypothesis }} {{eqn | r = \paren {k - 2} \times k^2 + 2 \times k^2 + k^2       | c =  }} {{eqn | o = >       | r = 1 + 2 \times k + k^2       | c =  }} {{eqn | r = \paren {k + 1}^2       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>3}: n! > n^2$ {{qed}}
We note that: {{begin-eqn}} {{eqn | l = 1!       | r = 1       | c =  }} {{eqn | r = 1^3       | c =  }} {{eqn | l = 2!       | r = 2       | c =  }} {{eqn | o = <       | r = 8       | c =  }} {{eqn | r = 2^3       | c =  }} {{eqn | l = 3!       | r = 6       | c =  }} {{eqn | o = <       | r = 27       | c =  }} {{eqn | r = 3^3       | c =  }} {{eqn | l = 4!       | r = 24       | c =  }} {{eqn | o = <       | r = 64       | c =  }} {{eqn | r = 4^3       | c =  }} {{eqn | l = 5!       | r = 120       | c =  }} {{eqn | o = <       | r = 125       | c =  }} {{eqn | r = 5^3       | c =  }} {{end-eqn}} The proof then proceeds by induction. For all $n \in \Z_{\ge 6}$, let $\map P n$ be the proposition: :$n! > n^3$ === Basis for the Induction === $\map P 6$ is the case: {{begin-eqn}} {{eqn | l = 6!       | r = 720       | c =  }} {{eqn | o = >       | r = 216       | c =  }} {{eqn | r = 6^3       | c =  }} {{end-eqn}} Thus $\map P 6$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 6$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k! > k^3$ from which it is to be shown that: :$\paren {k + 1}! > \paren {k + 1}^3$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \paren {k + 1}!       | r = \paren {k + 1} \times k!        | c =  }} {{eqn | o = >       | r = \paren {k + 1} \times k^3        | c = Induction Hypothesis }} {{eqn | r = \paren {k - 3} \times k^3 + 3 \times k^3 + k^3       | c =  }} {{eqn | o = >       | r = 1 + 3 \times k + 3 \times k^2 + k^3       | c =  }} {{eqn | r = \paren {k + 1}^3       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{>5}: n! > n^3$ {{qed}}
{{begin-eqn}} {{eqn	| l = \sum_{j \mathop = 1}^n j \times j! 	| r = \sum_{j \mathop = 1}^n \paren {j + 1 - 1} \times j! }} {{eqn	| r = \sum_{j \mathop = 1}^n \paren {\paren {j + 1} j! - j!} }} {{eqn	| r = \sum_{j \mathop = 1}^n \paren {\paren {j + 1}! - j!} }} {{eqn	| r = \paren {n + 1}! - 1! 	| c = Telescoping Series: Example 1 }} {{eqn	| r = \paren {n + 1}! - 1 }} {{end-eqn}} {{qed}}
The proof proceeds by induction on $n$. For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition: :$\dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$ First we investigate the edge case. Let $n = 1$. Then we have: {{begin-eqn}} {{eqn | l = \dbinom 1 0       | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | l = \dbinom 1 1       | r = 1       | c = Binomial Coefficient with Self }} {{end-eqn}} Thus we see: :there are no $k$ such that $0 \le k < \dfrac {1 - 1} 2 = 0$ and: :there are no $k$ such that $\dbinom 1 k < \dbinom 1 {k + 1}$ Thus $\map P 1$ is seen to hold. === Basis for the Induction === Let $n = 2$. Then we have: {{begin-eqn}} {{eqn | l = \dbinom 2 0       | r = 1       | c = Binomial Coefficient with Zero }} {{eqn | l = \dbinom 2 1       | r = 2       | c = Binomial Coefficient with One }} {{eqn | l = \dbinom 2 2       | r = 1       | c = Binomial Coefficient with Self }} {{end-eqn}} Thus we see: :there is one $k$ such that $0 \le k < \dfrac {2 - 1} 2 = \dfrac 1 2$, and that is $k = 0$ and: :$\dbinom 2 k < \dbinom 2 {k + 1}$ holds for exactly $k = 0$. Thus $\map P 2$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true. So this is the induction hypothesis: :$\dbinom r k < \dbinom r {k + 1} \iff 0 \le k < \dfrac {r - 1} 2$ from which it is to be shown that: :$\dbinom {r + 1} k < \dbinom {r + 1} {k + 1} \iff 0 \le k < \dfrac r 2$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \dbinom {r + 1} k       | r = \dbinom r k + \dbinom r {k - 1}       | c = Pascal's Rule }} {{eqn | o = <       | r = \dbinom r k + \dbinom r k       | rr= \iff 0 \le {k - 1} < \dfrac {r - 1} 2       | c = Induction Hypothesis }} {{eqn | o = <       | r = \dbinom r {k + 1} + \dbinom r k       | rr= \iff 0 \le {k - 1} < \dfrac {r - 1} 2 \text { and } 0 \le k < \dfrac {r - 1} 2       | c = Induction Hypothesis }} {{eqn | o = <       | r = \dbinom r {k + 1} + \dbinom r k       | rr= \iff 0 \le k < \dfrac {r - 1} 2       | c =  }} {{eqn | r = \dbinom {r + 1} {k + 1}       | rr= \iff 0 \le k < \dfrac {r - 1} 2       | c = Pascal's Rule }} {{end-eqn}} {{finish|Can't get my head round the inequalities on $k$, needs more thought}} So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{> 0}: \dbinom n k < \dbinom n {k + 1} \iff 0 \le k < \dfrac {n - 1} 2$
=== Sufficient Condition === Let $n$ be odd and $k = \dfrac {n - 1} 2$. Let $n = 2 m + 1$ for some $m \in \Z_{\ge 0}$. We have: {{begin-eqn}} {{eqn | l = k       | r = \dfrac {n - 1} 2       | c =  }} {{eqn | r = \dfrac {\paren {2 m + 1} - 1} 2       | c =  }} {{eqn | r = \dfrac {2 m} 2       | c =  }} {{eqn | r = m       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \dbinom n k       | r = \dbinom {2 m + 1} m       | c =  }} {{eqn | r = \dbinom {2 m + 1} {\paren {2 m + 1} - m}       | c = Symmetry Rule for Binomial Coefficients }} {{eqn | r = \dbinom {2 m + 1} {m + 1}       | c =  }} {{eqn | r = \dbinom n {k + 1}       | c =  }} {{end-eqn}} {{qed|lemma}} === Necessary Condition === Let $n$ and $k$ be such that: :$\dbinom n k = \dbinom n {k + 1}$ From Condition for Increasing Binomial Coefficients, it is not the case that: :$0 \le k < \dfrac {n - 1} 2$ as under such a condition we would have: :$\dbinom n k = \dbinom n {k + 1}$ So: :$k \ge \dfrac {n - 1} 2$ By the Symmetry Rule for Binomial Coefficients, we also have that: :$\dfrac {n - 1} 2 < k \le n \iff \dbinom n k > \dbinom n {k + 1}$ and so for $\dbinom n k = \dbinom n {k + 1}$ it is not the case that $\dfrac {n - 1} 2 < k \le n$. So: :$k \le \dfrac {n - 1} 2$ Hence that means: :$k = \dfrac {n - 1} 2$ which means: :$n = 2 k + 1$ and so $n$ is odd. {{qed}}
Write the differential equation as:  :$y' - k y = 0$ Taking Laplace transforms: :$\laptrans {y' - k y} = \laptrans 0$ From Laplace Transform of Constant Mapping, we have:  :$\laptrans 0 = 0$ We also have:  {{begin-eqn}} {{eqn	| l = \laptrans {y' - k y} 	| r = \laptrans {y'} - k \laptrans y 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = s \laptrans y - \map y 0 - k \laptrans y 	| c = Laplace Transform of Derivative }} {{end-eqn}} So: :$\paren {s - k} \laptrans y = \map y 0$ Giving:  :$\laptrans y = \dfrac {\map y 0} {s - k}$ So: {{begin-eqn}} {{eqn	| l = y 	| r = \map { {\mathcal L}^{-1} } {\frac {\map y 0} {s - k} } 	| c = {{Defof|Inverse Laplace Transform}} }} {{eqn	| r = \map y 0 \map { {\mathcal L}^{-1} } {\laptrans {e^{k x} } }  	| c = Linear Combination of Laplace Transforms, Laplace Transform of Exponential }} {{eqn	| r = \map y 0 e^{k x} 	| c = {{Defof|Inverse Laplace Transform}} }} {{end-eqn}} Setting $C = \map y 0$ gives the result. {{qed}}
{{begin-eqn}} {{eqn | l = \dfrac {\d y} {\d x}       | r = k y       | c =  }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} y       | r = \int k \rd x       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = \ln y       | r = k x + C'       | c = Primitive of Reciprocal, Primitive of Constant }} {{eqn | ll= \leadsto       | l = y       | r = e^{k x + C'}       | c =  }} {{eqn | r = e^{k x} e^{C'}       | c =  }} {{eqn | r = C e^{k x}       | c = putting $C = e^{C'}$ }} {{end-eqn}} {{qed}}
Let $\tilde C$ be any connected set such that: :$C \subseteq \tilde C$ Then $x \in \tilde C$. Hence $\tilde C \in \mathcal C_x$. From Set is Subset of Union,  :$\tilde C \subseteq C$. Hence $\tilde C = C$. It follows that $C$ is a maximal connected set of $T$ by definition.
By definition, $\tilde C \in \mathcal C_x$. From Set is Subset of Union, $\tilde C \subseteq C$. By maximality of $\tilde C$ then $\tilde C = C$
{{begin-eqn}} {{eqn | l = \dbinom n k       | r = \dbinom {n - 1} k + \dbinom {n - 1} {k - 1}       | c = Pascal's Rule }} {{eqn | r = \paren {\dbinom {n - 2} {k - 1} + \dbinom {n - 2} k} + \paren {\dbinom {n - 2} {k - 2} + \dbinom {n - 2} {k - 1} }       | c = Pascal's Rule (twice) }} {{eqn | r = \dbinom {n - 2} {k - 2} + 2 \dbinom {n - 2} {k - 1} + \dbinom {n - 2} k       | c = simplifying }} {{end-eqn}} In the expression $\dbinom {n - 2} {k - 2} + 2 \dbinom {n - 2} {k - 1} + \dbinom {n - 2} k$ we note that: :if $k < 2$ then $\dbinom {n - 2} {k - 2}$ has a negative coefficient on the bottom :if $k > n - 2$ then $\dbinom {n - 2} k$ has a coefficient on the bottom that is greater than $n$. Hence the usual comfortable range of $k$ is exceeded and so it cannot be guaranteed that the conditions are satisfied for the equation to be true. If $n \le 3$ then $2 \le k \le n - 2$ cannot be fulfilled. Hence the bounds on both $k$ and $n$. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$\displaystyle \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$ $\map P 0$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^0 2^j \binom n j       | r = \dbinom 0 0       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | r = 3^0       | c =  }} {{end-eqn}} Thus $\map P 0$ is seen to hold. === Basis for the Induction === $\map P 1$ is the case: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^1 2^j \binom n j       | r = \dbinom 1 0 + 2 \dbinom 1 1       | c =  }} {{eqn | r = 1 + 2       | c =  }} {{eqn | r = 3^1       | c =  }} {{end-eqn}} Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$\displaystyle \sum_{j \mathop = 0}^k 2^j \binom k j = 3^k$ from which it is to be shown that: :$\displaystyle \sum_{j \mathop = 0}^{k + 1} 2^j \binom {k + 1} j = 3^{k + 1}$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 0}^{k + 1} 2^j \binom {k + 1} j       | r = \binom {k + 1} 0 + \sum_{j \mathop = 1}^k 2^j \binom {k + 1} j + 2^{k + 1} \dbinom {k + 1} {k + 1}       | c = separating out top and bottom indices }} {{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom {k + 1} j + 2^{k + 1}       | c = Binomial Coefficient with Zero, Binomial Coefficient with Self }} {{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \paren {\binom k j + \binom k {j - 1} } + 2^{k + 1}       | c = Pascal's Rule }} {{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom k j + 2 \sum_{j \mathop = 1}^k 2^{j - 1} \binom k {j - 1} + 2^{k + 1}       | c =  }} {{eqn | r = 1 + \sum_{j \mathop = 1}^k 2^j \binom k j + 2 \sum_{j \mathop = 0}^{k - 1} 2^j \binom k j + 2^{k + 1}       | c = Translation of Index Variable of Summation }} {{eqn | r = 1 + \paren {\sum_{j \mathop = 0}^k 2^j \binom k j - 2^0 \binom k 0} + 2 \paren {\sum_{j \mathop = 0}^k 2^j \binom k j - 2^k \binom k k} + 2^{k + 1}       | c = rectifying the end points }} {{eqn | r = \sum_{j \mathop = 0}^k 2^j \binom k j + 2 \sum_{j \mathop = 0}^k 2^j \binom k j - 2 \times 2^k + 2^{k + 1}       | c = simplifying }} {{eqn | r = 3^k + 2 \times 3^k - 2 \times 2^k + 2^{k + 1}       | c = Induction Hypothesis }} {{eqn | r = 3^{k + 1}       | c = tidying up }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \Z_{\ge 0}: \displaystyle \sum_{j \mathop = 0}^n 2^j \binom n j = 3^n$ {{qed}}
{{begin-eqn}} {{eqn | l = 3^n       | r = \paren {2 + 1}^n }} {{eqn | r = \sum_{j \mathop = 0}^n 2^j 1^{n - j} \binom n j       | c = Binomial Theorem }} {{eqn | r =\sum_{j \mathop = 0}^n 2^j \binom n j }} {{end-eqn}} {{qed}}
Taking Laplace transforms:  :$\laptrans {y'' + 4 y} = \laptrans {8 x^2 - 4 x}$ We have:  {{begin-eqn}} {{eqn	| l = \laptrans {y'' + 4 y} 	| r = \laptrans {y''} + 4 \laptrans y 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = s^2 \laptrans y - s \map y 0 - \map {y'} 0 	| c = Laplace Transform of Second Derivative }} {{end-eqn}} We also have:  {{begin-eqn}} {{eqn	| l = \laptrans {8 x^2 - 4 x} 	| r = 8 \laptrans {x^2} - 4 \laptrans x 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = \frac {8 \cdot 2!} {s^3} - \frac {4} {s^2} 	| c = Laplace Transform of Power }} {{eqn	| r = \frac {16 - 4 s} {s^3} }} {{end-eqn}} So: :$\paren {s^2 + 4} \laptrans y = s \map y 0 + \map {y'} 0 + \dfrac {16 - 4 s} {s^3}$ Giving:  {{begin-eqn}} {{eqn	| l = \laptrans y 	| r = \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac {16 - 4 s} {s^3 \paren {s^2 + 4} } }} {{eqn	| r = \map y 0 \frac s {s^2 + 4} + \map {y'} 0 \frac 1 {s^2 + 4} + \frac 4 {s^3} + \frac {s + 1} {s^2 + 4} - \frac 1 {s^2} - \frac 1 s 	| c = partial fraction expansion }} {{eqn	| r = \paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s }} {{end-eqn}} So: {{begin-eqn}} {{eqn	| l = y 	| r = \map { {\mathcal L}^{-1} } {\paren {\map y 0 + 1} \frac s {s^2 + 4} + \paren {\map {y'} 0 + 1} \frac 1 {s^2 + 4} + \frac 4 {s^3} - \frac 1 {s^2} - \frac 1 s} 	| c = {{Defof|Inverse Laplace Transform}} }} {{eqn	| r = \paren {\map y 0 + 1} \map { {\mathcal L}^{-1} } {\frac s {s^2 + 2^2} } + \frac {\paren {\map {y'} 0 + 1} } 2 \map { {\mathcal L}^{-1} } {\frac 2 {s^2 + 2^2} } + 2 \map { {\mathcal L}^{-1} } {\frac {2!} {s^3} } - \map { {\mathcal L}^{-1} } {\frac {1!} {s^2} } - \map { {\mathcal L}^{-1} } {\frac {0!} s} 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = \paren {\map y 0 + 1} \map { {\mathcal L}^{-1} } {\laptrans {\cos 2 x} } + \frac {\paren {\map {y'} 0 + 1} } 2 \map { {\mathcal L}^{-1} } {\laptrans {\sin 2 x} } + 2 \map { {\mathcal L}^{-1} } {\laptrans {x^2} } - \map { {\mathcal L}^{-1} } {\laptrans x} - \map { {\mathcal L}^{-1} } {\laptrans 1} 	| c = Laplace Transform of Cosine, Laplace Transform of Sine, Laplace Transform of Power }} {{eqn	| r = \paren {\map y 0 + 1} \cos 2 x + \paren {\frac {\paren {\map {y'} 0 + 1} } 2 } \sin 2 x - 1 - x + 2 x^2 	| c = {{Defof|Inverse Laplace Transform}} }} {{end-eqn}} Setting $C_1 = \dfrac {\paren {\map {y'} 0 + 1} } 2$ and $C_2 = \map y 0 + 1$ gives the result. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 4$ :$\map R x = 8 x^2 - 4 x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$(2): \quad y'' + 4 y = 0$ From Linear Second Order ODE: $y'' + 4 y = 0$, this has the general solution: :$y_g = C_1 \sin 2 x + C_2 \cos 2 x$ From the Method of Undetermined Coefficients for Polynomials: :$y_p = A_0 + A_1 x + A_2 x^2$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A_0 + A_1 x + A_2 x^2       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = A_1 + 2 A_2 x       | c = Power Rule for Derivatives }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 2 A_2       | c = Power Rule for Derivatives }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = 2 A_2 + 4 \paren {A_0 + A_1 x + A_2 x^2}       | r = 8 x^2 - 4 x       | c =  }} {{eqn | ll= \leadsto       | l = 2 A_2 + 4 A_0       | r = 0       | c = equating coefficients }} {{eqn | l = 4 A_1       | r = -4       | c =  }} {{eqn | l = 4 A_2       | r = 8       | c =  }} {{eqn | ll= \leadsto       | l = A_1       | r = -1       | c =  }} {{eqn | l = A_2       | r = 2       | c =  }} {{eqn | l = A_0       | r = -1       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 \sin 2 x + C_2 \cos 2 x - 1 - x + 2 x^2$ is the general solution to $(1)$. {{qed}}
Taking Laplace transforms:  :$\laptrans {y'' - y' - 6 y} = \laptrans {e^{-x} }$ We have:  {{begin-eqn}} {{eqn	| l = \laptrans {y'' - y' - 6 y} 	| r = \laptrans {y''} - \laptrans {y'} - 6 \laptrans y 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = s^2 \laptrans y - s \map y 0 - \map {y'} 0 - \paren {s \laptrans y - \map y 0} - 6 \laptrans y }} {{eqn	| r = \paren {s^2 - s - 6} \laptrans y - \paren {s \map y 0 + \map {y'} 0 - \map y 0} }} {{end-eqn}} We also have:  {{begin-eqn}} {{eqn	| l = \laptrans {e^{-x} } 	| r = \frac 1 {s - \paren {-1} }  	| c = Laplace Transform of Exponential }} {{eqn	| r = \frac 1 {s + 1} }} {{end-eqn}} So: :$\paren {s^2 - s - 6} \laptrans y = s \map y 0 + \paren {\map {y'} 0 - \map y 0} + \dfrac 1 {s + 1}$ Giving:  {{begin-eqn}} {{eqn	| l = \laptrans y 	| r = \map y 0 \paren {\frac s {s^2 - s - 6} } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {s^2 - s - 6} } + \frac 1 {\paren {s + 1} \paren {s^2 - s - 6} } }} {{eqn	| r = \map y 0 \paren {\frac s {\paren {s - 3} \paren {s + 2} } } + \paren {\map {y'} 0 - \map y 0} \paren {\frac 1 {\paren {s - 3} \paren {s + 2} } } + \frac 1 {\paren {s + 1} \paren {s - 3} \paren {s + 2} } 	| c = factorising }} {{eqn	| r = \frac {\map y 0} 5 \paren {\frac 2 {s + 2} + \frac 3 {s - 3} } + \frac {\map {y'} 0 - \map y 0} 5 \paren {\frac 1 {s - 3} - \frac 1 {s + 2} } + \frac 1 {20} \paren {\frac 1 {s - 3} + \frac 4 {s + 2} - \frac 5 {s + 1} } 	| c = partial fraction expansion }} {{eqn	| r = \paren {\frac {2 \map y 0 - \map {y'} 0 + \map y 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {3 \map y 0 + \map {y'} 0 - \map y 0} 5 + \frac 1 {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }  }} {{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }  }} {{end-eqn}} So:  {{begin-eqn}} {{eqn	| l = y 	| r = \map { {\mathcal L}^{-1} } {\paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \frac 1 {s + 2} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \frac 1 {s - 3} - \frac 1 {4 \paren {s + 1} }  } 	| c = {{Defof|Inverse Laplace Transform}} }} {{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \map { {\mathcal L}^{-1} } {\frac 1 {s + 2} } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \map { {\mathcal L}^{-1} } {\frac 1 {s - 3} } - \frac 1 4 \map { {\mathcal L}^{-1} } {\frac 1 {s + 1} } 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} \map { {\mathcal L}^{-1} } {\laptrans {e^{-2 x} } } + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } \map { {\mathcal L}^{-1} } {\laptrans {e^{3 x} } } - \frac 1 4 \map { {\mathcal L}^{-1} } {\laptrans {e^{-x} } } 	| c = Laplace Transform of Exponential }} {{eqn	| r = \paren {\frac {3 \map y 0 - \map {y'} 0 + 1} 5} e^{-2 x} + \paren {\frac {8 \map y 0 + 4 \map {y'} 0 + 1} {20} } e^{3 x} - \frac 1 4 e^{-x} 	| c = {{Defof|Inverse Laplace Transform}} }} {{end-eqn}} Setting:  :$C_1 = \dfrac {8 \map y 0 + 4 \map {y'} 0 + 1} {20}$ :$C_2 = \dfrac {3 \map y 0 - \map {y'} 0 + 1} 5$ gives the result. {{qed}}