$$\frac{dx}{dy} = \frac{x(\alpha - \beta y)}{y(\delta x - \gamma)}$$    How do I extract two differential equations (y as a function of x and x as a function of y) from the equation above? I could separate the variables, but I don't see how that would help.
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.
98449
$M(x,y)dx + N(x,y)dy=0$ is said to be a perfect differential when   $$\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}$$  Let $M_y=\frac{\partial (M(x,y))}{\partial y}$ and $N_x=\frac{\partial (N(x,y))}{\partial x}$.  In case if:  $\frac{M_y-N_x}{N(x,y)}$ is only a function of x only (say $f(x)$) then it has a integrating factor $e^{\int f(x)dx}$.  But if $\frac{M_y-N_x}{N(x,y)}$ is a constant then what will be the integrating factor?     Say for this problem:   $(axy^2+by)dx+(bx^2y+ax)dy=0$   Is there any other method to solve this differential equation?
91199
I asked 3 professors at my university and none gave me a clear cut answer, but instead merely told me qualities of this notation. Here is what I understand so far from what they told me:   1)Treat the top variable as as variable when finding the derivative 2)Treat the bottom variable as a constant when finding derivative 3)It it said "Find x with respect to y", but what exactly does that mean? What does it mean for something to be in respect to something else?  It seems like $\frac{dx}{dy}$ notation changes according to values in the problem. For instance, If $y = x^3 + 2x$ and $\frac{dx}{dt} = 5$, find $\frac{dy}{dt}$ when $x=2$. Why do the values of $\frac{dx}{dy}$ change in this problem and how do I solve 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?
70885
I've run into a strange situation while trying to apply Integration By Parts, and I can't seem to come up with an explanation.  I start with the following equation:  $$\int \frac{1}{f} \frac{df}{dx} dx$$  I let:  $$u = \frac{1}{f} \text{ and } dv = \frac{df}{dx} dx$$  Then I find:  $$du = -\frac{1}{f^2} \frac{df}{dx} dx \text{ and } v = f$$  I can then substitute into the usual IBP formula:  $$\int udv = uv - \int v du$$  $$\int \frac{1}{f} \frac{df}{dx} dx = \frac{1}{f} f - \int f \left(-\frac{1}{f^2} \frac{df}{dx}\right) dx$$  $$\int \frac{1}{f} \frac{df}{dx} dx = 1 + \int \frac{1}{f} \frac{df}{dx} dx$$  Then subtracting the integral from both sides, I've now shown that:  $$0 = 1$$  Obviously there must be a problem in my derivation here...  What wrong assumption have I made, or what error have I made?  I'm baffled.
76345
Let $f(x,y)$ be a function of two variables. Given that  $$\frac{\partial f}{\partial x}=0$$ for all $y$, and  $$\frac{\partial f}{\partial y}=0$$ for all $x$, is there a theorem that states $f(x,y)=$ constant for all $x$, $y$? Or how would I prove this rigorously?
94839
A friend told me to solve the following differential equation:  $$f'(x)=f(f(x))$$  I have no idea how to solve this! This doesn't seem to be an ordinary differential equation and I can't even solve this numerically!  I think my friend is trolling me.
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.
55612
let $\alpha'(x)=\beta(x), \beta'(x)=\alpha(x)$ and assume that $\alpha^2 - \beta^2 = 1$. how would I go about calculating the following anti derivative : $\int (\alpha (x))^5 (\beta(x))^4$d$x$.  I have tried many different methods with no success. Firstly, I tried to write the integral in terms of one of the variables (using the first condition) and then using integration by parts. This resulted in a very complex answer which I abandoned midway. Additionally, I have tried to use an appropriate U-substitution which unfortunately does not seem to work. This is because when I take the first derivative of my u - due to chain rule - I get a product with one variable being either $\alpha'(x)$ or $\beta'(x)$ and as a result I cannot continue further with the integration due to not being able to get the entire integral in terms of only u. Any help would be appreciated. Thank you.
57224
Solve the differential equation: $yy'^2-2xy'+y=0$ (*)  I have   $(*)\Leftrightarrow (yy')^2-2xyy'+y^2=0\\ \Leftrightarrow z'^2-4xz+4z^2 =0 \ \ (z=y^2)$  I can't know how to continue.  Could you give me some hints. Thanks for helping.
80595
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$   My Attempt $$ x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\ 2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\ \frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\ \frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy} $$ How do I proceed further and find the derivative ?
39920
$$f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} &amp; \text{if } (x,y) \neq (0,0), \\ 0 &amp; \text{if } (x,y) = (0,0). \end{cases}$$ I tried finding both mixed partial derivatives but they ended up being the same for that function. I must be failing to take into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this?
41609
I read the following statement in a book on Calculus, as part of my mathematics course:  Technically this separation of $\frac{dy}{dx}$ is not mathematically valid. However, the resulting integration leads to correct answer.  The book also contains the following:  To solve a differential equation by separation of variables:  get all the $x$ values on one side and all the $y$ values on the other side by multiplication and division. separate $\frac{dy}{dx}$ as if it were a fraction. integrate both sides.   Note: This box doesn't refer to a particular problem. It refers to a class of problems of differential equations which can be solved using the Method of Separation of Variables. My high school mathematics teacher told me that this is the most fundamental way to solve differential equations but the textbook says it is not mathematically valid. I am not able to understand why are certain methods being followed without having a mathematical proof. Or am I wrong?
47708
I am looking to find out how the second derivative formula works.  I can blindly apply it but I don't have a grasp of what is going on or why. The first derivitive has the formula $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. I told this was because if you divide them you cancel out the dts and get dy/dx. The second derivative has the formula $\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{ \frac{d}{dt}\left(\frac{dy}{dx}\right) }{\frac{dx}{dt}}$. Is this something that can be explained with only a knowledge of calculus or is this something I would have to take on faith until I learn more. Any help would be greatly appreciated.
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 $f = 0$?
28945
I am trying to solve the problem:     \begin{array}{l}{\text { A particular solution of the differential equation }} \\ {x \frac{\mathrm{d} v}{\mathrm{d} x}=f(v) \text { has } v=2 \text { when } x=\mathrm{e} \text { . }} \\ {\text { If } k \text { is the value of } v \text { when } x=1, \text { show that } \int_{k}^{2} \frac{1}{f(y)} \mathrm{d} y=1}\end{array}   I realize that: $\int\frac{dv}{f(v)} = \int \frac{dx}{x}$. I am unsure of how to proceed.
111094
Could you help me to solve equation  $$y' = \exp \left(-\frac yx\right) + \frac yx;\quad y(e) = 0$$  I know how to solve 1st order linear de like $y' = \exp \bigl(-\frac 1x\bigr) + \frac yx$ but here I have the dependent variable in the part that usually (in my practice) was free of it.