Learn First Order Problems with These Differential Equations Examples

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.

I have a simple, two object thermodynamic model with radiation and advection. This model consists of two first order quadratic differential equations, what I would like to solve analytically. The equations can be simplified to
$\frac{d}{dt}x=a_0{x}^4+a_1{y}^4+a_{2}x+a_{3}y+a_4$
$\frac{d}{dt}y=a_5{x}^4+a_6{y}^4+a_{7}x+a_{8}y+a_9$
I'm trying to find the analytic solution because I would like to use the model in an embedded system with limited resources, where solving it numerically is not feasible. So far I have tried to solve this with Maxima without any success. I have also consulted with my college textbooks, but they cover only linear differential equation systems.

Consider the linear first order non-homogeneous partial differential equation
$U_x+y{U}_y-<!--\; -\; -->y=y{e}^{-<!--\; -\; -->x}$
By using the method of characteristics show that its general solution is given by
$u(x,y)=-<!--\; -\; -->y{e}^{-<!--\; -\; -->x}+e^{x}g(y{e}^{-<!--\; -\; -->x})$
where g is any differentiable function of its argument
$$\begin{gathered} A=1,B=y,d=-1,f=y{e}^{-x} \\ {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y \\ dy=ydx \\ \int <!--\; \int \; -->{\displaystyle \frac{1}{y}}dy=\int <!--\; \int \; -->dx \\ \ln <!--\; \; -->y=x+c \\ \mathrm{\eta <!--\; \eta \; -->}(x,y)=c \\ \text{let }\mathrm{\xi <!--\; \xi \; -->}=x \end{gathered}$$
Then
$\mathrm{\eta <!--\; \eta \; -->}=\ln <!--\; \; -->y-<!--\; -\; -->x$
$\mathrm{\xi <!--\; \xi \; -->}=x$
those are my characteristics curves,
when I put it into $[a{\mathrm{\xi <!--\; \xi \; -->}}_x+b{\mathrm{\xi <!--\; \xi \; -->}}_y]{\mathrm{\omega <!--\; \omega \; -->}}_{\mathrm{\xi <!--\; \xi \; -->}}+d\mathrm{\omega <!--\; \omega \; -->}=F$
I get the equation ${\mathrm{\omega <!--\; \omega \; -->}}_{\mathrm{\xi <!--\; \xi \; -->}}-<!--\; -\; -->\mathrm{\omega <!--\; \omega \; -->}=e^{\mathrm{\eta <!--\; \eta \; -->}}$
Is this all correct?

I was solving for the equation of a curve and I arrived at the following differential equation. However I do not know how to solve it:
$\frac{dy}{dx}=\frac{-<!--\; -\; -->x+\sqrt{x^2+y^2}}{y}$

I have this simple differential equation: $y^{\prime }=(\tan <!--\; \; -->x)y.$
after integrating $\frac{y^{\prime }}{y(x)}=\tan <!--\; \; -->x$
i came up with $\log <!--\; \; -->y(x)=-<!--\; -\; -->\log <!--\; \; -->\cos <!--\; \; -->(x)+1.$
now my question is this one: why $e^{-<!--\; -\; -->\log <!--\; \; -->\cos <!--\; \; -->(x)}=\sec <!--\; \; -->(x)?$

The differential equation that describes my system is given as
$y^{(n)}+a_{n-<!--\; -\; -->1}{y}^{(n-<!--\; -\; -->1)}+\cdots <!--\; \cdots \; -->+a_1\stackrel{\dot{}<!--\; \dot{}\; -->}{y}+a_{0}y=b_{n-<!--\; -\; -->1}{u}^{(n-<!--\; -\; -->1)}+\cdots <!--\; \cdots \; -->+b_1\stackrel{\dot{}<!--\; \dot{}\; -->}{u}+b_{0}u+g(y(t),u(t))$
I want to express the above differential equation into a system of linear differential equations of the form
$\stackrel{\dot{}<!--\; \dot{}\; -->}{x}=Ax+Bu+B_{p}g$
$y=Cx$
The matrices are given as follows: However, I am not able to prove, how to get them
$A=\left[\begin{array}{ccccc}0& 1& 0& \cdots <!--\; \cdots \; -->& 0\\ 0& 0& 1& \cdots <!--\; \cdots \; -->& 0\\ ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->& ⋮<!--\; ⋮\; -->\\ -<!--\; -\; -->a_0& -<!--\; -\; -->a_1& -<!--\; -\; -->a_2& \cdots <!--\; \cdots \; -->& -<!--\; -\; -->a_{n-<!--\; -\; -->1}\end{array}\right]$
$C=\left[\begin{array}{ccccc}1& b_1/b_0& b_2/b_0& \cdots <!--\; \cdots \; -->& b_{n-<!--\; -\; -->1}/b_0\end{array}\right]$
How do I get the above matrices from the differential equation form as shown above?

$y^{‴}(t)+sin(t)y^{″}(t)-<!--\; -\; -->g(t)y^{\prime }(t)+g(t)y(t)=f(t)$
Write the following third order differential equation as an equivalent system of first order ordinary differential equations and write it in the form
$\frac{dz}{dt}=f(t,z)$
Here is what I have so far:
$x_1(t)=y(t)$
$x_2(t)=y^{\prime }(t)$
$x_3(t)=y^{″}(t)$
so I can then do,
$x_1^{\prime }(t)=y^{\prime }(t)$
$x_2^{\prime }(t)=y^{″}(t)$
$x_3^{\prime }(t)=y^{‴}(t)=f(t)-<!--\; -\; -->sin(t)y^{″}(t)+g(t)y^{\prime }(t)-<!--\; -\; -->g(t)y(t)$
$x_3^{\prime }(t)=f(t)-<!--\; -\; -->sin(t)x_3(t)+g(t)x_2(t)-<!--\; -\; -->g(t)x_1(t)$
and I think the dynamics function is:
$f(t,z)=[x_1^{\prime }(t),x_2^{″}(t),x_3^{‴}(t)]$
$f(t,z)=[x_2(t),x_3(t),f(t)-<!--\; -\; -->sin(t)x_3(t)+g(t)x_2(t)-<!--\; -\; -->g(t)x_1(t)]$
I'm not sure if I'm doing this correctly. What does difference does the f(t) and g(t) make in this case? what is z exactly?

I have been trying to solve a differential equation as a practice question for my test, but I am just unable to get the correct answer. Please have a look at the D.E:
$dy/dx=1/(3x+\sin <!--\; \; -->(3y))$
My working is as follows:
$dx/dy=3x+\sin <!--\; \; -->(3y)$
$dx-<!--\; -\; -->3x=\sin <!--\; \; -->(3y)dy$
Integrating both sides:
$x-<!--\; -\; -->(3/2)x^2=-<!--\; -\; -->(1/3)\cos <!--\; \; -->(3y)+c$
But the correct answer is:
$x=c{e}^{3}y-<!--\; -\; -->1/6(\cos <!--\; \; -->(3y)+\sin <!--\; \; -->(3y))$, which is quite different from what I have got. Could someone please help me solve this?

I'm looking at the following problem. I'm trying to solve for y, and separating variables doesn't look like an option. I'm thinking putting it into standard form of a first order linear eq would be a good place to start. Is this correct?
$\frac{y^{\prime }-<!--\; -\; -->e^{-<!--\; -\; -->t}+3}{y}=-<!--\; -\; -->3$
$y(0)=5$
In standard form:
$y^{\prime }+P(x)y=Q(x)$
$y^{\prime }+3y=-<!--\; -\; -->3+e^{-<!--\; -\; -->t}$
$y{e}^{3t}=\int <!--\; \int \; -->-<!--\; -\; -->3e^{3t}+e^{2t}\ast <!--\; \ast \; -->dt$
$y{e}^{3t}=e^{-<!--\; -\; -->3t}+0.5e^{2t}+C$
$y=-<!--\; -\; -->1+0.5e^{-<!--\; -\; -->t}+C$
Solving for initial condition y(0) = 5:
$y=-<!--\; -\; -->1+0.5e^{-<!--\; -\; -->t}+C$
$y(0)=-<!--\; -\; -->1+0.5e^0+C=5$
$y(0)=-<!--\; -\; -->0.5+C=5$
$y(0)=C=5.5$
Plug back in C:
$y=-<!--\; -\; -->1+0.5e^{-<!--\; -\; -->t}+5.5$
I've went through the steps of calculating the integrating factor and solving for y. The answer was incorrect. Is this the right way to approach this problem?