$$y'' -4y' + 4y=0$$  My attempt:  We solve $r^2e^{rx}-4re^{rx}+e^{rx}=0$ for $r$, this gives $r=2$  Then the solution is: $y=c_1e^{2x}$  But the solution given is $y=c_1e^{2x}+c_2xe^{2x}$.  I wonder where the $x$ in $c_2xe^{2x}$ comes from, why is it different?
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.
27935
I can't seem to find the  solutions to  $z^4+1=0 $. $z$ is in the complex plane.   The solutions show four roots; however, how do I find them once $z^4 = -1$?
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?
101912
How to solve differential equation $$y=xy'+\dfrac12y'^2$$
17689
When solving the differential equation $y'' + ay' + by = 0$ (with constant, real coefficients $a$ and $b$, although they could be complex if you like), you do it by setting up the characteristic equation $r^2 + ar + b = 0$, finding its solutions $r_1, r_2$, and then the general solution to this equation is $Ce^{r_1x} + De^{r_2x}$. This works both when the solutions are real and when they are complex.  However, when we have a double root $r_1 = r_2$, we get a different general solution, namely $Ce^{r_1x} + Dxe^{r_1x}$. I have no trouble seing that this is indeed a solution, and intuitive reasoning on degrees of freedom dictates that we must have a linear combination of two terms in our general solution, while $e^{r_1x}$ and $e^{r_2x}$ are the same. So the fact that there is a second term of some other form is not surprising.  I have, however, yet to see a "natural" explanation of this $xe^{r_1x}$ term. If one were developing the theory from scratch, how would one find this solution (other than blind luck)? If I wanted to teach ODE's to a class of students "the right way", i.e. with good explanations and motivations for everything (as opposed to just pulling out ready-made solutions like what was done to me when I was learning this exact thing), how would I motivate even considering a term like $xe^{r_1x}$ (other than "Well, exponentials aren't quite cutting it, but this is kindof like an exponential, right? Let's try it.")? And is there a way of solving the general differential equation that does not involve splitting into cases depending on whether the characteristic polynomial is a square?
16282
How to solve this equation $$x^{2}=2^{x}$$   where $x \in \mathbb{R}$.  Por tentativa erro consegui descobri que $2$ é uma solução, mas não encontrei um método pra isso. Alguma sugestão?(*)  (Translation: By trying different values I've found that $2$ is a solution, but I couldn't find any method to this though. Any suggestions? )
85965
Consider the following differential equation with constant coefficients: $$ y'' + 2ky' + k^2y = 0 \, . $$ The auxiliary equation is \begin{align} m^2 + 2km + k^2 &amp;= 0 \\[4pt] (m+k)^2 &amp;= 0 \\[4pt] m &amp;= -k \, . \end{align} This means that the solutions of the differential equation are $Ae^{-kx}+B\color{red}{x}e^{-kx}$, where $A$ and $B$ are arbitrary constants. The $x$-term highlighted in red is what confuses me. Generally speaking, if the roots of the auxiliary equation are $\alpha$ and $\beta$, then the solutions to the differential equation are $Ae^{\alpha x} + Be^{\beta x}$. However, if $\alpha=\beta$, then the solutions are not $Ae^{\alpha x} + Be^{\alpha x}$—or at least, this solution is incomplete. I can verify that $y=xe^{-kx}$ is a solution of the differential equation by plugging it back in to the equation, but I'm looking for a deeper reason for why this occurs.
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.
57706
I am asked to solve $$u_x+u_y=1$$ If is was homogeneous i.e., $u_x+u_y$ the answer would be $u(x,y)=f(y-x)$ where $f$ is an arbitrary function. I have found the following set of solutions: $$u(x,y)=\lambda x +(1-\lambda)y$$ where $\lambda$ is an arbitrary constant(real or imaginary). I just have no idea what method other then trial and error would have lead me here. Any ideas? Thanks!
99252
I have equations \begin{align} 7z'(x) &amp;= x^2 + y(x)^2 + z(x)^2  \\ y''(x) &amp;= -(y'(x) + 7y(x))\sin(z(x)) \end{align} where $$(y(0),y'(0),z(0))=(1.5,-2.6,0.5)$$  The task is to solve this problem using a computer. Can someone help with explaining what goes in $u'(x)$?  Thanks.  I am guessing $u(x) = (y'',z')$.
28243
If $x,y \in \mathbb Z$ , then find all the solutions of   $$y^2=3x^4+3x^2+1$$       I was asked this question by my friend who said that he encountered this while solving another problem. I have tried several things but am unable to solve this question. Moreover, this has to be done using elementary methods only. So far, I have tried to factorize and use Pell's equation. At the end, I'm getting $$2y_{n} + (2x^2_{n}+1)\sqrt{3}=(2+\sqrt{3})^{n}$$   where $n \in \mathbb Z^{+}$    But I'm not able to figure out how to show a contradiction from here. Can anyone please help me out? Thanks.
39021
The question is to investigate the number of zeroes of $y''(x)+e^{x^2}y(x)=0$ in $[0,3π]$.  Solving this ODE would not be an easy task as one has to use the power series solution and then investigating the zeroes of the solution will require more analysis. I thought it to compare this ODE with the standard $y''(x)+y(x)=0$ whose solution has three or four zeroes in the interval $[0,3π]$.  Since the coefficient of $y(x)$ is $e^{x^2}\ge 1$ for $x\in [0,3π]$ so the solution of given ODE must have atleast three zeroes in $[0,3π]$. However, what  I thought was in the lights of Sturm-Comparison theorem so I am not sure.  Am I correct to interpret this?
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?
76180
I have been following the thread " Convert Airy's Equation $y''-xy=0$  to Bessel equation $$t^2u''+tu'+(t^2-c^2)u$$ " but I can't join the dots to a solve similar equation $y''+x^2y=0$ so as to obtain a solution of the form $$\sqrt x\left(AJ_\frac{1}{4}+BJ_{-\frac{1}{4}}\right)$$ I actually get an equation that looks this way $$t^2\frac{du}{dt}+t\frac{du}{dt}+(t^2+\frac{5}{64})u$$ The above equation can not yield the desired solution. Please help me to clearly see this. Thank you.
58995
Find all possible solutions of $$2^x + 3^y = z^2.$$    My approach.   First I substituted $x = 0$, and got the solution, then for $y = 0$. And for $x &gt;  0$ and $y &gt; 0$ , I just know the Pythagorean triple and got the solution as $(4,2,5)$ and $(4,2,-5)$. Please help me to solve it properly.
7879
I'm trying to solve this equation:  $$2-x=-\sqrt{x}$$ Multiply by $(-1)$: $$\sqrt{x}=x-2$$ power of $2$: $$x=\left(x-2\right)^2$$ then: $$x^2-5x+4=0$$ and that means: $$x=1, x=4$$    But $x=1$ is not a correct solution to the original equation.  Why have I got it? I've never got a wrong solution to an equation before. What is so special here?