necessaryh
2021-03-11
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Nathalie Redfern

Answered 2021-03-12
Author has **99** answers

This is a separable equation, which can be written as
$(2y+\mathrm{cos}y)dy=6{x}^{2}dx$

Integrate:$\int (2y+\mathrm{cos}y)dy=\int 6{x}^{2}dx$

LHS:$\int (2y+\mathrm{cos}y)dy={y}^{2}+\mathrm{sin}y+C1,$
where C1 is some constant.

RHS:$\int 6{x}^{2}dx=2{x}^{3}+C2$
where C2 is some constant.

So, from (1) it follows that${y}^{2}+\mathrm{sin}y=2{x}^{3}+C$
where we defined C as C2-C1

Integrate:

LHS:

RHS:

So, from (1) it follows that

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I found some transformed tables, but I couldn't understand where the result comes from, or how to reproduce the accounts step by step.

$${\mathcal{L}}^{-1}\frac{1}{(s+\lambda {)}^{2}-{\omega}^{2}}$$

$${\mathcal{L}}^{-1}\frac{a(s+2\lambda )+b}{(s+\lambda {)}^{2}-{\omega}^{2}}$$

I found some transformed tables, but I couldn't understand where the result comes from, or how to reproduce the accounts step by step.