# y'=6x^2/(2y+cosy))

Question
Laplace transform
$$\displaystyle{y}'={6}\frac{{x}^{{2}}}{{{2}{y}+{\cos{{y}}}}}{)}$$

2021-03-12
This is a separable equation, which can be written as $$\displaystyle{\left({2}{y}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}={6}{x}^{{2}}{\left.{d}{x}\right.}$$
Integrate: $$\displaystyle∫{\left({2}{y}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}=∫{6}{x}^{{2}}{\left.{d}{x}\right.}$$
LHS: $$\displaystyle∫{\left({2}{y}+{\cos{{y}}}\right)}{\left.{d}{y}\right.}={y}^{{2}}+{\sin{{y}}}+{C}{1},$$ where C1 is some constant.
RHS: $$\displaystyle∫{6}{x}^{{2}}{\left.{d}{x}\right.}={2}{x}^{{3}}+{C}{2}$$ where C2 is some constant.
So, from (1) it follows that $$\displaystyle{y}^{{2}}+{\sin{{y}}}={2}{x}^{{3}}+{C}$$ where we defined C as C2-C1

### Relevant Questions

Solve differential equation $$\frac{\cos^2y}{4x+2}dy= \frac{(\cos y+\sin y)^2}{\sqrt{x^2+x+3}}dx$$

Use Laplace transform to solve the folowing initial value problem $$y"+2y'+2y=0$$
$$y(0)=2$$
$$y'(0)=-1$$
Use Laplace transform to solve the following initial-value problem
$$y"+2y'+y=0$$
$$y(0)=1, y'(0)=1$$
a) \displaystyle{e}^{{-{t}}}+{t}{e}^{{-{t}}}\)
b) \displaystyle{e}^{t}+{2}{t}{e}^{t}\)
c) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{t}\)
d) \displaystyle{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
e) \displaystyle{2}{e}^{{-{t}}}+{2}{t}{e}^{{-{t}}}\)
f) Non of the above
Use Laplace transform to find the solution of the IVP
$$2y'+y=0 , y(0)=-3$$
a) $$f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}$$
b)$$f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}$$
c)$$f{{\left({t}\right)}}={6}{e}^{{{2}{t}}} d) \(f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}$$
ALSO, USE PARTIAL FRACTION WHEN YOU ARRIVE
$$L(y) = \left[\frac{w}{(s^2 + a^2)(s^2+w^2)}\right]*b$$
Problem 2 Solve the differential equation
$$\frac{d^2y}{dt^2}+a^2y=b \sin(\omega t)$$ where $$y(0)=0$$
and $$y'(0)=0$$
use the Laplace transform to solve the initial value problem.
$$y"-3y'+2y=\begin{cases}0&0\leq t<1\\1&1\leq t<2\\ -1&t\geq2\end{cases}$$
$$y(0)=-3$$
$$y'(0)=1$$
Find the solutions for the given linear systems of differential equations using Laplace Transforms.
$$w'-2y'+3w=0: y"+w=2\sin x$$
$$y(0)=w(0)=2 , y’(0)=-1$$
Use the Laplace transform to solve the given system of differential equations.
$$\frac{(d^2x)}{(dt^2)}+\frac{(d^2y)}{(dt^2)}=\frac{t}{2}$$
$$\frac{(d^2x)}{(dt^2)}-\frac{(d^2y)}{(dt^2)}=4t$$
$$x(0) = 5, x'(0) = 0,$$
$$y(0) = 0, y'(0) = 0$$
$$y'''-2y"-21y'-18y=-18$$
$$y(0)=2$$
$$y'(0)=7$$
$$y"(0)=95$$
Use Laplace transform to solve the folowing initial value problem $$y"+2y'+y=4e^{-t} y(0)=2 y'(0)=-1$$