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You can, as alternative solution method, avoid using a partial fraction decomposition by treating the ODE as a Bernoulli DE $$x'=3x^2-15x$$ so that by setting $u=x^{-1}$ you get $$u'=-x^{-2}x'=15x^{-1}-3=15u-3$$ which now is a linear first order DE in $u$ and additionally separable. Integration leads to $$5u(t)-1=e^{15t}(5u(0)-1)\implies 5\frac{x(0)}{x(... 1 Another way would be to explicitly calculate the coefficients of the partial fractions. We are given that$$\frac{dx}{dt}=3x(x-5)$$which rearranges to (observing that x\equiv 0 and x \equiv 5 are solutions to the ODE but not the IVP)$$\frac{1}{x(x-5)}dx=3dt \implies-\frac{1}{5}\Big(\frac{(x-5)-x}{x(x-5)}\Big)dx=3dt\implies -\frac{1}{5}\Big(\frac{1}{...

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HINT \begin{align*} \frac{1}{3x(x-5)} = \frac{1}{3}\times\frac{1}{x(x-5)} = \frac{1}{15}\times\frac{x - (x-5)}{x(x-5)} = \frac{1}{15(x-5)} - \frac{1}{15x} \end{align*}

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Before giving you a complete solution I want to point out the mistakes in your solution. Your partial fractions are correct. As mentioned in the comments $\int \frac{1}{ax} dx \ne \ln | a x |$. Instead we have $$\int \frac{1}{ax} dx = \frac{1}{a} \int \frac{1}{x} dx = \frac{1}{a} \ln | x|$$ for $a \ne 0$. We have $e^{\ln(a) - \ln(b)} \ne a - b$. This is ...

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$$1 = \frac{A}{3x} + \frac{B}{x-5}$$ if and only if $$(3x)(x-5) = A(x-5) + B(3x) \text{,}$$ with $x \not\in \{0,5\}$. How do we get "$15xB$"?

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You ask for a generating function. So define $A(z) = \sum_{n \ge 0} a_n z^n$, shift your recurrence (no subtraction in indices is nicer). Then multiply by $z^n$ and sum over $n \ge 0$, recognize some sums: \begin{align*} a_{n + 1} &= a_n + 5 n^2 + 4 n - 5 \\ \sum_{n \ge 0} a_{n + 1} z^n &= \sum_{n \ge 0} a_n z^n + 5 \sum_{... 1 $$\frac{x^2+1}{x^3-x}=\frac{x^2-1+2}{x^3-x}=\frac{1}{x}+\frac{2}{x(x-1)(x+1)}=$$ $$=\frac{1}{x}+\frac{2}{x+1}\left(\frac{1}{x-1}-\frac{1}{x}\right)=\frac{1}{x}+\frac{1}{x-1}-\frac{1}{x+1}+2\left(\frac{1}{x+1}-\frac{1}{x}\right)=$$ $$=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}{x-1}.$$ Can you end it now? 7 Hint: sincex^3-x=x(x-1)(x+1)$you can easily split$f(x)$into partial fractions. 15 Write your function as $$\frac{x^2 + 1}{x^3 - x} = \frac{1}{x-1} + \frac{1}{x+1} - \frac{1}{x}$$ 1 Let$A,B$be unknown constants. Let$S=\{x\in\mathbb{R}\mid x\ne\pm 1\}$. Then for all$x\in S, \begin{align*} &\frac{1}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}\\[4pt] \implies\;&1=A(x+1)+B(x-1)\\[4pt] \implies\;&A(x+1)+B(x-1)-1=0\\[4pt] \end{align*} Now consider the polynomial functionp:\mathbb{R}\to\mathbb{R}given by $$p(x)=A(x+1)+B(x-1)-1$$ ... 0 Let's take your example. We have \begin{align}\frac{1}{n(n+1)(n+2)} = \frac{a}{n} + \frac{b}{n+1} + \frac{c}{n+2}&=\color{red}{\boxed{\frac an+\frac{b(n+2)+c(n+1)}{(n+1)(n+2)}\cdot\frac nn}}\quad(\text{group together}\,b,c)\\&=\color{blue}{\boxed{\frac b{n+1}+\frac{a(n+2)+cn}{n(n+2)}\cdot\frac{n+1}{n+1}}}\quad(\text{group together}\,a,c)\\&=\... 2 This is a nice shortcut. It uses the observation that the function on the LHS has divergences at the isolated values ofn$that make the denominator zero, and each term on the RHS must match one of those divergences (in both location and magnitude). So, suppose in general that the LHS is $$\frac{1}{(n-z_1)(n-z_2)\ldots(n-z_k)}=\frac{c_1}{n-z_1} + \frac{... 7 I'll just start with a as an example. Multiply both sides by n to get$$\frac 1{(n+1)(n+2)} = a + \frac {bn}{n+1} + \frac {cn}{n+2}$$Since this is true for all n, plug in 0. The b and c terms become 0, leaving a in the formula described above:$$\frac 1{(0+1)(0+2)} = a$$14 Just multiply the whole formula by the linear factor that you are setting to zero. It cancels on the left side and in exactly one term on the right side. in your second example with the factor n+2 this gives$$ \frac1{n(n+1)}=(n+2)\frac an+(n+2)\frac{b}{n+1}+c $$Then set the linear factor to zero, that is, evaluate at its root and , giving exactly the ... 2 Let X=x-1 then$$\frac{3x^2-8x+6}{x(x-1)^2(2-x)}=-\frac{3(X+1)^2-8(X+1)+6}{(X+1)X^2(X-1)}= -\frac{3X^2-2X+1}{X^2(X^2-1)}$$Now you can use the first part. 0 By your work the coefficient before x it's 1, which ends. Also,$$\frac{x+6}{x^2-x-6}=\frac{x+6}{(x-3)(x+2)}=\frac{x-3+9}{(x-3)(x+2)}==\frac{1}{x+2}+\frac{9}{5}\left(\frac{1}{x-3}-\frac{1}{x+2}\right)=\frac{\frac{9}{5}}{x+3}+\frac{-\frac{4}{5}}{x+2},$$which gives A+B=1. 3 Put x=-2 and x=3 to get B=-\frac{4}{5} and A=\frac{9}{5} respectively. 2 If x+6=(x+2)A+B(x-3) then$$x+6=x(A+B)+(2A-3B)\impliesA+B=1\text{ and } 2A-3B=6$$2 By equating coefficients, you have the system of equations$$A+B = 1,2A-3B = 6.$$0 There is no need to do partial fractions on this one. Simply set$\frac{1}{t}=x$. Your new integral will be of the form$\frac{t}{t^2-1}\$ and that's simply a natural log. That's it! Can you finish from here?

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