Consider the following model. Find the general solution    $\frac{dx}{dt}=-2y$, $\frac{dy}{dt}=8x$  So far this is what I have:   $$\frac{dx}{dt}=-2y$$ $$x'=-2y$$ $$y=-\frac{x'}{2}$$ $$y=-\frac{x''}{2}$$ and then $$\frac{dy}{dt}=8x$$ $$y'=8x$$ $$-\frac{x''}{2}=8x$$ $$-x''=16x$$ $$x''+16x=0$$  Is this correct so far? What step should I take next?
123806
Solve the following system:   $$dx/dt=-.2(y-2)$$ $$dy/dt=.8(x-2)$$  This is what I have so far, but I got stuck.. $$\begin{eqnarray} dx/dt&amp;=&amp;-.2y-.4\\ x'&amp;=&amp;-.2y-.4\\ x'+.4&amp;=&amp;-.2y\\ y&amp;=&amp;-x/2'-2\\ y&amp;=&amp;-x''/2-2\\ \\ dy/dt&amp;=&amp;.8(x-2)\\ dy/dt&amp;=&amp;.8x-1.6\\ y'&amp;=&amp;.8x-1.6\\ -x''-2&amp;=&amp;.8x-1.6\\ -x''&amp;=&amp;.8x+.4\\ x''+16x+.8&amp;=&amp;0 \end{eqnarray}$$  when I used the quadratic formula I got stuck. Is this correct so far? Can someone finish it off for me?
17331
I am not sure whether this question is even well-posed. But today I learnt that $e^Df(x) = f(x+1)$ where $D$ is differential operator and  $$e^D \triangleq \sum_{i=0}^{\infty} \frac{D^i}{i!}.$$  (ref. )  So I was curious as to whether the differential equation  $$\frac{df(t)}{dt} = e^Df(t) = f(t+1)$$  has any solutions apart from $f = 0$?
88316
Consider the ODE $$\frac{dy}{dx}=x^{2}+y^{2}, y(0)=0.$$ I have to find the interval of unique solution by using Picard method. As in my local book it is solved as $$|f(x,y)|=|x^{2}+y^{2}|\leq a^{2}+b^{2}=M. $$ $$h=\min\{a,\frac{b}{M}\}$$ where $M=a^{2}+b^{2}$. So $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$. Then my main problem is the author do like $$a=\frac{b}{a^{2}+b^{2}}$$ which gives a quadratic in $b$ as $ab^{2}-b+a^{3}=0$ with discriminant as  $1-4a^{4}=0$ gives $a=\frac{1}{\sqrt{2}}$ so $|h|\leq\frac{1}{\sqrt{2}}.$ Please suggest me how he  find $h=\min\{a,\frac{b}{a^{2}+b^{2}}\}$? Why he put $a=\frac{b}{a^{2}+b^{2}}$? Thanks in advance.
121843
$$\frac{dx(t)}{dy(t)}=\frac{\alpha x(t) - \beta x(t) y(t)}{-\gamma y(t) + \delta x(t)y(t)}$$  How would one simplify this fraction? Maybe the chain rule could be of any use, but I don't see how.
78760
Express the 2nd order ODE $$\begin{align}\mathrm d_t^2 u:=\frac{\mathrm d^2 u}{\mathrm dt^2}&amp;=\sin(u)+\cos(\omega t)\qquad \omega \in \mathbb Z /\{0\} \\u(0)&amp;=a\\\mathrm d_t u(0)&amp;=b\end{align}$$ as a system of 1st order ODEs and verify there exists a global solution by invoking the global existence and uniqueness theorems.  I'm not sure how to express second order ODEs as first order ODEs, any tips?
103425
How should I approach this problem? Should I put ln logarithm on both sides and move factor of variable of lnx on right side in front of the ln, or do somehing else? I understand general solving and types of solving these differential equations of first order, but I have no idea what my next step should be.
92873
Please guide me:  $y' + ay +b = 0$ ($a$ not zero) is supposed to be separable and has solution  $y = ce^{-ax} - \frac ba$  Here is my start to this problem:  $\frac{dy}{dx} + ay = -b$ is as far as I can go with this. How should I go about separating $x$ and
71858
$y\frac{\text{d}^2y}{\text{d}x^2}-2\left(\frac{\text{d}y}{\text{d}x}\right)^2+xy^3\frac{\text{d}y}{\text{d}x}=0$. I want to get solution in elementary functions in order to do asymptotic analysis. I tried the problem with Mathematica. It did not give any solutions.
115725
I am able to solve simple differential equations like :  $$\dfrac{dy}{dx} = 3x^2 + 2x$$  We simply bring $dx$ to other site and integrate.  But how do we find solutions of differential equations like :  $$\frac{d^2x}{dt^2} + \dfrac{kx}{m} = 0$$  ?  We have been told the solution is $x(t) = A\cos ( \omega t + \phi_o)$ where $\omega\ =\sqrt{\dfrac{k}{m}}$  but how do we actually find it?
37516
I am looking for help on how to find the solution of the following differential equation,  $$x\ddot{x}-\dot{x}^2=1,$$  which comes from solving the Euler-Lagrange equations for the lagrangian $L=x\sqrt{\dot{x}^2+1}$. It is from an exercise in my field theory course. My professor provided me with the solution $x(t)=c_1\cosh{\left(\frac{t-c_2}{c_1}\right)}$, which is a solution when plugged into the equation, but does not give any derivation.   Is there a way to solve this DE other than an educated guess? If so how?
9909
Let $A,B \in M_n(\mathbb{R})$, and suppose $[A,[A,B]] = [B,[A,B]] = 0$, where $[A,B] = BA-AB$. We want to show $$ e^{At}e^{Bt}=e^{(A+B)t}e^{[A,B]t^2/2}, \quad \forall t\in \mathbb{R}. $$ I'm given a hint to show that $x(t) =e^{-(A+B)t}e^{Bt}e^{At}x_0$ is a solution to the ODE $x'= t[A,B]x$ for each $x_0\in \mathbb{R}$.    I haven't been able to prove the hint, let alone see why being a solution helps prove the identity. Taking the derivative of $x(t)$ didn't lead to much, and moreover there's no $t$ outside of the exponent, and substituting it into the RHS of the ODE also didn't help.    The RHS of the identity to be shown solves $x' = (A+B+t[A,B])x$, which maybe gives some intuition about the ODE in the hint, but I'm not really sure where to go.
109377
I would like to solve:  $(\frac{d^{2}y}{dx^{2}})^2+\frac{d^{3}y}{dx^{3}} \frac{dy}{dx}=0$   Thanks in advance.
22316
solve the differential equation.   $$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$  The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.   Thanks in advance.
86097
The question is basically to find out the solution of the differential equation $$(2x^2+3y^2-7)xdx-(3x^2+2y^2-8)ydy=0$$.  Since the given differential equation is non homogeneous I tried to find out the point of intersection of $2x^2+3y^2-7=0$ and $3x^2+2y^2-8=0$.I got the point of intersection and then I tried to shift the coordinate system to that point.this removed the constant terms.However I was not able to proceed further.Please help me in this regard.thanks.
102774
I am studying for an exam about ODEs and I am struggling with one of the past exam questions. The past exam shows one exercise which asks us to solve: $$y' = \frac{1}{2} a y^2 + b y - 1$$with $y(0)=0$  The solution is given as $$y(x) = \frac{2 \left( e^{\Gamma x} - 1 \right)}{(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma},$$ with $\Gamma = \sqrt{b^2 + 2a}$    I am really getting stuck at this exercise and would love to have someone show me how this solution is derived.  One thing I did find out is that this ODE is variable separable. That is, $$y' = g(x)h(y) = (1) \cdot (\frac{1}{2} ay^2 + by - 1),$$ and therefore the solution would result from solving $$\int \frac{1}{\frac{1}{2} ay^2 + by - 1} dy = \int dx + C,$$ where C is clearly zero because $y(0) = 0$.  I am now getting stuck at solving the left integral. Could anyone please show me the steps?    UPDATE  So I came quite far with @LutzL solution, however my answers seems to slightly deviate from the solution given above. These are the steps I performed (continuing from @LutzL's answer):  You complete the square $\frac12ay^2+by-1=\frac12a(y+\frac ba)^2-1-\frac{b^2}{2a}$ and use this to inspire the change of coordinates $u=ay+b$ leading to $$ \int \frac{dy}{\frac12ay^2+by-1}=\int\frac{2\,du}{u^2-2a-b^2} $$ and for that your integral tables should give a form using the inverse hyperbolic tangent. Or you perform a partial fraction decomposition for $$ \frac{2Γ}{u^2-Γ^2}=-\frac{1}{u+Γ}+\frac{1}{u-Γ} $$ and find the corresponding logarithmic anti-derivatives, $$ \ln|u-Γ|-\ln|u+Γ|=Γx+c,\\ \frac{u-Γ}{u+Γ}=Ce^{Γx},\ C=\pm e^c $$ which you now can easily solve for $u$ and then $y$.  Given that $y(0) = 0$ we have $u(y(0)) = u(0) = b$ and therefore the final equation becomes $$\frac{u(0)-Γ}{u(0)+Γ}= \frac{b-Γ}{b+Γ}=Ce^{Γ\cdot0} = Ce^{Γ \cdot 0} = C$$ Now by first isolating $u$ I get $$u - \Gamma = u C e^{\Gamma x} + \Gamma C e^{\Gamma x} \Rightarrow \\ u \left( 1 - C e^{ \Gamma x} \right) = \Gamma \left( 1 + C e^{\Gamma x} \right) \Rightarrow \\ u = \frac{\Gamma \left( 1 + C e^{\Gamma x} \right)}{\left( 1 - C e^{ \Gamma x} \right)}$$ Now substituting u and C gives $$ay + b= \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right)}{\left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)} \Rightarrow \\ y = \frac{\Gamma \left( 1 + \frac{b-Γ}{b+Γ} e^{\Gamma x} \right) - b \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}{a \left( 1 - \frac{b-Γ}{b+Γ} e^{ \Gamma x} \right)}$$  Now using the fact that $\Gamma = \sqrt{b^2 + 2a} \Rightarrow a = \frac{(\Gamma + b)(\Gamma - b)}{2}$ we get that $$y = \frac{2 \left( \Gamma \left( 1 + \frac{b-\Gamma }{b+\Gamma } e^{\Gamma x} \right) - b \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right) \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)}  \\ = \frac{2 \left( (\Gamma - b) + (\Gamma + b) \frac{b-\Gamma }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)} \\ = \frac{2 \left( (\Gamma - b) - (\Gamma + b) \frac{\Gamma - b }{b+ \Gamma } e^{\Gamma x} \right)}{(\Gamma + b)(\Gamma - b) \left( 1 - \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} \right)}$$ Now cancelling the terms $(b + \Gamma)$ and $(b - \Gamma)$ wherever possible and multiplying denominator and nominator by -1 gives $$y = \frac{2 \left(  e^{\Gamma x} - 1 \right)}{(\Gamma + b) \left( \frac{b-\Gamma }{b+\Gamma } e^{ \Gamma x} - 1 \right)} $$  So clearly, I got the nominator right, but I can not seem to get the denominator to equal $(b + \Gamma)(e^{\Gamma x} - 1) + 2\Gamma$. Can someone rescue me and show me what I did wrong?  Maybe it helps if I say that x is always positive?
120773
[1] $y'' - 4y' + 4y = 0$  Usually problem like these will have the answer in the form $C_1e^a + C_2e^b ... $ where $a $ and $b$ are the roots of the characteristic equation $e^{rt}$  $$ y = e^{rt} $$ $$ y' = re^{rt}.. y'' = r^2 e^{rt} $$  $$ r^2e^{rt} - 4e^{rt} + 4e^{rt} = 0$$ $$ e^{rt}(r-2)(r-2) = 0$$ $$ y= C_1e^{2t} + C_2e^{2t}$$  However, this is not correct as the answer is $ y = C_1e^{2t} + C_2te^{2t}$ ! I just don't know why.  [2] For a similar problem , $y'' + 3y' - 4y = 0 $ I did the exact same thing and the answer is $$y = c_1e^t + c_2e^{-4t} $$  How are [1] and [2] different? They look the same, why does [2]'s solution have an extra factor of t.
95252
$$ Ae^{Bx} + Cx = D $$  Solve for $x$, where $A, B, C, D$ are real constants. This popped up in a run of the mill high school first-order DE question. No matter what I try with logarithms it doesn't seem like I can isolate an $x$ onto one side of the equation. Is this equation unsolvable analytically? If not, how does one generally go about solving something like this?