Solve System of Differential Equations with Expert Help

I am aware that the solution to a homogeneous first order differential equation of the form $\frac{dy}{dx}=p(x)y$ can be obtained by simply by rearranging to:
$\frac{dy}{y}=p(x)dx$
Then it is simply a question of integrating both sides and the answer is straightforward. Now what would happen if RHS had a constant, how a can find a particular solution to this case: $\frac{dy}{dx}=p(x)y+C$
I know that the general solution would be the sum of the homogeneous equation and the particular solution

Transform the differential equation
$\{\begin{array}{ll}{u}^{‴}(t)=\sin <!--\; \; -->(u^{″}(t))-<!--\; -\; -->u^2(t),& t>0,\\ u^{(i)}=u_i\text{for }i\in <!--\; \in \; -->\{0,1,2\}\end{array}$
into an equivalent system of first order ordinary differential equations.
I know how to carry out this procedure for linear systems but I am stuck because of the $\sin <!--\; \; -->(u^{″}(t))$ term.
For far I tried setting $x:\equiv <!--\; \equiv \; -->u^{″}$ and $y:\equiv <!--\; \equiv \; -->u^{\prime }$ and then substituting into the equation to obtain multiple differential equations in x and y but don't know how to simplify them to obtain the desired result:
$\{\begin{array}{l}{y}^{\prime }(t)=x(t),\\ x^{\prime }(t)=\sin <!--\; \; -->(x(t))-<!--\; -\; -->u^2(t),\\ y^{″}(t)=\sin <!--\; \; -->(y^{\prime }(t))-<!--\; -\; -->u^2(t).\end{array}$
For context: This is the first part of a two part question where in the second part one is asked to find the iterative rule for Eulers method.

Problem:
Solve the following differential equation:
$\begin{array}{rcl}6x^{2}ydx-<!--\; -\; -->(x^3+1)dy& =& 0\end{array}$
Answer:
This is a separable differential equation.
$\begin{array}{rcl}\frac{6x^2}{x^3+1}dx-<!--\; -\; -->\frac{dy}{y}& =& 0\\ \int <!--\; \int \; -->\frac{6x^2}{x^3+1}dx-<!--\; -\; -->\int <!--\; \int \; -->\frac{dy}{y}& =& c_1\\ 2\ln <!--\; \; -->|x^3+1|-<!--\; -\; -->\ln <!--\; \; -->|y|& =& c_1\\ \ln <!--\; \; -->(x^3+1{)}^2-<!--\; -\; -->\ln <!--\; \; -->|y|& =& c_1\\ \ln <!--\; \; -->(\frac{(x^3+1{)}^2}{|y|})& =& c_1\\ (x^3+1{)}^2& =& c|y|\end{array}$
However, the book gets:
$\begin{array}{rcl}(x^3+1{)}^2& =& |cy|\end{array}$
Is my answer different from the book's answer? I believe it is.

I'm taking an elementary differential equations class and we got our first exam back and I don't understand why I was wrong on one of the questions. I got a 96, and the professor said there were quite a few d's and f's, so I didn't want to quibble about this.
Anyway, the problem was ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{1}{x+y+2}}$ and I chose the sub $u=x+y+2$, with ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{dy}{du}}{\displaystyle \frac{du}{dx}}={\displaystyle \frac{dy}{du}}(1+{\displaystyle \frac{1}{u}})$ and then plugged in the expression for ${\displaystyle \frac{dy}{dx}}$, separated the equation and solved for y in terms of u and then substituted back to get $y=\ln <!--\; \; -->(x+y+3)+c.$ The professor took issue with my expression for ${\displaystyle \frac{dy}{dx}}$, saying that I couldn't differentiate y w.r.t. u if y is part of u. Is this correct?

My question is
If u(x,y) and v(x,y) are two integrating factors of a diff eqn M(x,y)dx+N(x,y)dy, u/v is not a constant. then u(x,y)=cv(x,y)is a solution to the differential eqn for every constant c. I m totally stuck :(
Another doubt i have is how to derive the singular solution for the Clairaut's equation. i tried it we have y=px+f(p) diff wrt x and considering dp/dx=0 we get p=c, how to solve the other part?

I have the following exepression in my book:
$\frac{dx}{dt}+a_1(t)x=g(t),x(t_0)=x_0$
Then it says, multiply both sides of the differential equation by the integrating factor $I(t)$.
$I(t)\frac{dx(t)}{dt}+a_1(t)I(t)x(t)=I(t)g(t)$
So far so good. Hereafter it says, the left-hand side is an exact derivative.
$\frac{d[x(t)I(t)]}{dt}=I(t)g(t)$
And my question is, how does the book come to the last? Can anyone give a HINT.

If a particle of mass $m$ moves in the $x$-$y$ plane, then its equations of motion are
$m\frac{d^{2}x}{d{t}^2}=f(t,x,y)\text{and}m\frac{d^{2}y}{d{t}^2}=g(t,x,y).$
Here $f$ and $g$ represent the $x$ and $y$ components, respectively, of the force acting on the particle. Replace this system of two second-order equations by an equivalent system of four first order equations of the form:
$y_1^{\prime }=f_1(x,y_1,...,y_n)$
$y_2^{\prime }=f_2(x,y_1,...,y_n)$
$y_n^{\prime }=f_n(x,y_1,...,y_n)$

I understand how replace a differential equation by an equivalent system of first order equations when the differential looks something like
$x{y}^{″}-<!--\; -\; -->x^2{y}^{\prime }-<!--\; -\; -->x^{3}y=0$
An equivalent system is:
$y_0^{\prime }=y_1$
$y_1^{\prime }=x^2{y}_0+x{y}_1$
Therefore, I need someone to point me in the right direction for my question stated at the top.

How to solve the differential equation $(dy/dx{)}^2=(x-<!--\; -\; -->y{)}^2$ with initial condition $y(0)=0$?
I solved the equation by partitioning it into two differential equations.
1) $dy/dx=x-<!--\; -\; -->y$ The solution is $\to <!--\; \to \; -->$ $1-<!--\; -\; -->x+y=-<!--\; -\; -->\exp <!--\; \; -->(-<!--\; -\; -->x)$
and
2) $1+x-<!--\; -\; -->y=\exp <!--\; \; -->(x)$
How do we write combined solution of such equations.

The problem is of the following form: $\frac{dr}{dt}$=$f(r)$, so $\frac{1}{f(r)}dr=dt$. My goal is to get some r(t) from this differential equation by numerically integrating this, that is, a value for r at every t. The conditions are $r(0)=r_0$ and $r(\mathrm{\infty <!--\; \infty \; -->})=\mathrm{\infty <!--\; \infty \; -->}$. The problem I run into is that I am not quite sure what the limits would be to get such values for r(t). Any suggestions on where to go from this would be greatly appreciated, thanks.

Express the differential equation
$y^{‴}-<!--\; -\; -->6y^{″}-<!--\; -\; -->y^{\prime }+6y=0$
as a system of first order equations i.e. a matrix equation of the form
$A(\stackrel{\to <!--\; \backslash rightarrow\; -->}{x}{)}^{\prime }=0$
where
$\stackrel{\to <!--\; \backslash rightarrow\; -->}{x}\text{ is the vector }\left[\begin{array}{r}{x}_1\\ x_2\\ x_3\end{array}\right].$

How can I find an exact solution for this problem ? Is there any technique for cubic nonlinearity as in the case of Bernoulli differential equation?
$y^{\prime }=x^3{y}^3-<!--\; -\; -->1$

I've found this particular equation rather tough, can you give me some hints on how to solve
$\stackrel{\dot{}<!--\; \dot{}\; -->}{y}+t\cos <!--\; \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}y}{2}+1-<!--\; -\; -->t=y$

Given the following differential equation i have to find the general solution using the integrating factor technique.
$\frac{dy}{dx}+\frac{y}{x}=e^x$
I know that P(x)=$\frac{1}{x}$ and Q(x)=$e^x$ also i found $\mathrm{\mu <!--\; \mu \; -->}(x)=e^{lnx}$
Using the formula
$y=\frac{1}{\mathrm{\mu <!--\; \mu \; -->}(x)}\int <!--\; \int \; -->\mathrm{\mu <!--\; \mu \; -->}(x)Q(x)dx$
i found this general solution:
$\frac{e^{x}x}{x}-<!--\; -\; -->\frac{e^x}{x}+\frac{C}{x}$
Help, maybe this is incorrect

I have the following differential equation
$s(1-<!--\; -\; -->s)t=f^{\prime }(t)(f(t)-<!--\; -\; -->st)$
Initial condition: $f(0)=0.$
I solved it and got the solution as
$f(t)=\sqrt{2c_1+s^2-<!--\; -\; -->2(s-<!--\; -\; -->1)s\ln <!--\; \; -->(t)}+s$
But the answer given is
$f(t)=k(s)t,$
where
$k(s)=\frac{s+\sqrt{4s-<!--\; -\; -->3s^2}}{2}.$
If anyone can provide me some hint on how to proceed and reach the specified answer, I would be really grateful.

I'm trying to understand what is wrong with this solution, since I'm not getting the same answer in Matlab
$y^{\prime }-<!--\; -\; -->xy=x{y}^{3/2},y(1)=4$
$\begin{array}{rlr}{y}^{\prime }-<!--\; -\; -->xy=& x{y}^{3/2}& \\ {\displaystyle \frac{dy}{dx}}=& x(y+y^{3/2})& \\ {\displaystyle \frac{dy}{(y+y^{3/2})}}=& xdx& \\ -<!--\; -\; -->2\ln <!--\; \; -->{\displaystyle \frac{1+\sqrt{y}}{\sqrt{y}}}=& {\displaystyle \frac{x^2}{2}}+c& \\ y=& {\displaystyle \frac{-<!--\; -\; -->1}{(1-<!--\; -\; -->{\mathrm{e}}^{\frac{-<!--\; -\; -->x^2}{4}+c}{)}^2}}& \\ 4=& {\displaystyle \frac{-<!--\; -\; -->1}{(1-<!--\; -\; -->{\mathrm{e}}^{\frac{-<!--\; -\; -->1}{4}+c}{)}^2}}& \\ c=& -<!--\; -\; -->0.655& \\ y=& {\displaystyle \frac{-<!--\; -\; -->1}{(1-<!--\; -\; -->{\mathrm{e}}^{\frac{-<!--\; -\; -->x^2}{4}-<!--\; -\; -->0.655}{)}^2}}& \end{array}$
This is what I'm getting in Matlab
$\left(\begin{array}{c}\frac{{(\text{tanh}(\frac{x^2}{8}+\text{atanh}\left(3\right)-<!--\; -\; -->\frac{1}{8})+1)}^2}{4}\\ \frac{{(\text{tanh}(-<!--\; -\; -->\frac{x^2}{8}+\text{atanh}\left(5\right)+\frac{1}{8})-<!--\; -\; -->1)}^2}{4}\end{array}\right)$

So I have the differential equation
$(y^2+xy)dx+(3xy+x^2)dy=0$. I am asked to solve this ODE by using the substitution $v=\frac{y}{x}$ but I run into some issues if I do that.
If $v=\frac{y}{x}$ then $y^{\prime }(x)=v+x{v}^{\prime }$
I then divided every term by $x^2$ to get
$(\frac{y^2}{x^2}+\frac{y}{x})+(\frac{3y}{x}+1)(\frac{dy}{dx})=0$
If I substitute, I then get,
$(v^2+v)+(3v+1)(v+x{v}^{\prime })=0$
Now I run into the problem where I can't make this into a separable equation...
I know that this is an exact equation and I can solve it if I use an integration factor $u(x)=y$ but I am asked to solve this question using the substitution $v=\frac{y}{x}$. Even if I muliply each term by the integration factor, things become even worse...
Any guidance would be appreciated. Thanks.

How would I solve this differential equation for $y(x)$?
$\frac{dy}{dx}=\frac{y-<!--\; -\; -->xy}{x-<!--\; -\; -->xy}$
$y-<!--\; -\; -->\ln <!--\; \; -->(y)=x-<!--\; -\; -->\ln <!--\; \; -->(x)+C$
I'm not sure what to do at this point. I looked it up on WolframAlpha and the solution uses something called a Product Log Function. How does it work? And how does the solution come out to be:
$y(x)=-<!--\; -\; -->W(-<!--\; -\; -->e^{(c_1-<!--\; -\; -->x)}x)$

Problem 1.7 in G.Teschl ODE and Dynamical Systems asks me to transform the following differential equation into autonomous first-order system:
$\stackrel{¨<!--\; ¨\; -->}{x}=t\sin <!--\; \; -->(\stackrel{\dot{}<!--\; \dot{}\; -->}{x})+x$
Transforming the ODE to a system is in this case easy, but whats the usual technique to transform it to an AUTONOMOUS system?

I need to solve an equation of this type
${\left(\frac{dy}{dx}\right)}^2+\frac{y^2}{b^2}=a^2$
but I don't know how to start Any help would be welcome, Thanks !

I am trying to solve a first order differential equation with the condition that $g(y)=0$ if $y=0$:
$\begin{array}{rl}& a{g}^{\prime }(cy)+b{g}^{\prime }(ey)=\mathrm{\alpha <!--\; \alpha \; -->}\\ \text{(1)}& & g(0)=0,\end{array}$
where parameters a,b,c,e are real nonzero constants; $\mathrm{\alpha <!--\; \alpha \; -->}$ is a complex constant; function $g(y):\mathbb{R}\to <!--\; \backslash rightarrow\; -->\mathbb{C}$ is a function mapping from real number y to a complex number. The goal is to solve for function $g(\cdot <!--\; \cdot \; -->)$. This is what I have done. Solve this differential equation by integrating with respect to y:
$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\mathrm{\alpha <!--\; \alpha \; -->}y+\mathrm{\beta <!--\; \beta \; -->},\end{array}$
where $\mathrm{\beta <!--\; \beta \; -->}$ is another complex constant. Plugging in y=0 and using the fact that g(0)=0, we have $\mathrm{\beta <!--\; \beta \; -->}=0$. Therefore, we have
$\begin{array}{r}\frac{a}{c}g(cy)+\frac{b}{e}g(ey)=\mathrm{\alpha <!--\; \alpha \; -->}y.\end{array}$
The background of this problem is Cauchy functional equation, so my conjecture is one solution could be $g(y)=\mathrm{\gamma <!--\; \gamma \; -->}y$. Plugging in $g(y)=\mathrm{\gamma <!--\; \gamma \; -->}y$, I get $\mathrm{\gamma <!--\; \gamma \; -->}=\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{a+b}$, which implies that one solution is $g(y)=\frac{\mathrm{\alpha <!--\; \alpha \; -->}}{a+b}y$. Then, I move on to show uniqueness. I define a vector-valued function $h\equiv <!--\; \equiv \; -->(h_1,h_2{)}^T$ such that
$\begin{array}{rl}{h}_1(y)& =\frac{a}{c}{g}_1(cy)+\frac{b}{e}{g}_1(ey)\\ h_2(y)& =\frac{a}{c}{g}_2(cy)+\frac{b}{e}{g}_2(ey),\end{array}$
where $g(y)\equiv <!--\; \equiv \; -->g_1(y)+i{g}_2(y)$. Then, I rewrite this differential equation as
$\begin{array}{rl}& h^{\prime }(y)=\mathrm{\alpha <!--\; \alpha \; -->}\\ \text{(2)}& & h(0)=0,\end{array}$
where $\mathrm{\alpha <!--\; \alpha \; -->}\equiv <!--\; \equiv \; -->({\mathrm{\alpha <!--\; \alpha \; -->}}_1,i{\mathrm{\alpha <!--\; \alpha \; -->}}_2{)}^T$. By the uniqueness theorem of first order differential equation, solution h(y) is unique. I have two questions. First, I think equation (1) and (2) should be equivalent. However, it seems that equation (1) can imply equation (2) but equation (2) may not imply equation (1). This is because h(0)=0 may imply either $g_1(0)=0,g_2(0)=0$ or $\frac{a}{c}+\frac{b}{e}=0$. Second, I have only proved that h(y) is unique. How should I proceed to show g(y) is also unique.