I'm not really sure on how to go about solving this problem...  Consider the initial value problem.       $$y' (x) = \frac{1}{2}(-x + \sqrt{x^2 + 4y})$$   $$y(2) = -1$$   $a)$ Show that $y(x) = 1-x$ and $y(x) = -\frac{x^2}{4}$ are two solutions to the above IVP.
79212
Prove that the solutions $y_1$ and $y_2$ for the initial value problem   $y'=$${-x+\sqrt{(x^2+4y)}}\over 2$ , $y(2)=-1$  are:  $y_1=1-x$  $y_2={{-x^2}\over 4}$  And explain why the two solutions not contradiction with the theory of existence and uniqueness.  ( sorry I don't speak English well )  My answer:  For the initial condition :  $y_1=1-x$ Satisfy the initial condition  $y(2)=-1$  $y_1=1-2=-1$   And Satisfy the differential equation  ${y_1}^{\prime}=-1$  ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4(1-x)}}\over 2}$  $={{-x+\sqrt{(x-2)^2}}\over 2}$  $={{-x+x-2}\over 2}=-1$  L.h.s.=R.h.s.  $y^{\prime}={{-x+\sqrt{x^2+4y}}\over 2}$  Also $y_2$ Satisfy the initial condition  $y_2={{-x^2}\over4}={-4\over4}=-1$   And satisfy the differential equation.   $y_2^{\prime}={-x\over 2}$  ${{-x+\sqrt{x^2+4y}}\over 2}={{-x+\sqrt{x^2+4({-x^2\over 4})}}\over 2}={-x\over 2}$  And for second part of exercise:  ${\partial f\over \partial y}={1\over \sqrt{x^2+4y}}$  and this not defined when $x=2, \ y=-1$  So $f(x,y)$ no satisfy lipshtize condition in any rectangular has (2,-1), Lipshtize condition including uniqueness of solution, so no contradict with the existence and uniqueness.  True ?  If this also wrong I well be delet it , I answered by the same way of my teacher :(
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Find the extremals of the functional    $$\text{J}(y)=  y^2(1)  + \int_0^1 y'^2(x)dx   , \ \ y(0)=1.$$   Answer: $y(x)=1-\frac{1}{2}x$   My solution:   $ F (x,y,y')=y'^2(x)$  After solving the Euler Lagrange equation we get $\frac{\mathrm{d}}{\mathrm{d}x}(2y')=0$  Which implies that $y=\frac{a}{2}x+b$ , using initial value condition we get $b=1$. Could you please help me find value of $a$?
48060
Let   $$y''-q(x)y=0$$be a differential equation with initial conditions  on $0\leq x&lt;\infty,$ as $y(0)=1,y'(0)=1$ where $q(x)$ is a positive monotonically increasing continuous function. Then which of the following are true?   $y(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y'(x)\rightarrow\infty$ as $x\rightarrow\infty$. $y(x)$ has finitely many zeros in $[0,\infty)$. $y(x)$ has infinitely many zeros in $[0,\infty)$.   Please don't mind, actually I am new in differential equation. I only know that by Picard's theorem the above differential equation has the unique solution, but I don't know what is the solution as I tried by direct hit and trial method. According to me the solution of above differential equation will be some thing in exponential form, so according to me its answer will be $a$, $b$, $c$. But I don't know the exact method. Please help me to solve the above problem. Thanks in advance.
42094
Just had a quick question regarding the uniqueness of the general solution of first-order differential equations:  For $x'(t)=Ax(t)$ where $A$ is an $n\times n$ matrix. and initial condition $x(0)=x_0$, the unique solution is $x(t)=e^{tA}x_0$.  The proof for showing that the solution is unique involves differentiating the matrix product $e^{-tA}x(t)$.  Could someone please explain why we always multiply by $e^{-tA}$.   Many thanks, Best.
22316
solve the differential equation.   $$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$  The question is from IIT entrance exam paper. I have tried substituting $x^2=t\ and \ y^2=u$ but was not a worth try.   Thanks in advance.
63158
I have not idea how to approach this differential equation.   $$y'-\int_0^xy(t)dt=2$$.  Basically, I did,  $$F''(t)-F(x)+F(0)=2 \;\;\;\;\;\;\; F'=y$$  I am stuck.  Thank You.
102569
Find the solution of following differential equation:     $$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin (y^2)+1)}$$   Could someone hint me something to get through this problem?
84034
I need help solving this past exam question, the professor posted exam questions with out solutions to last year's exam, I tried reduction of order, however that seemed to complicate things. And then I tried Euler-Cauchy method however I realized that won't work due to the $(2x^2 + 1)y''$.  Given $y_1=x$ is a solution $$(2x^2 + 1)y'' − 4xy' + 4y = 0$$ Find the second solution $y_2$ for equation.
110927
I have:  $\dfrac {dx} {dt}$=$-x+y$  $\dfrac {dy}{dt}$=$-x-y$  and I am trying to find $x(t)$ and $y(t)$ given that $x(0)=0$ and $y(0)=1$  I know to do this I need to decouple the equations so that I only have to deal with one variable but the decoupling is what I am having trouble with  Do I set them equal to each other and then just move like terms to separate sides getting two different equations and then integrate?
85885
If $x,y\in \mathbb{R}$ and $f(x+y)=f(x)+f(y)+y\sqrt{f(x)}$ and $f'(0)=0\;,$ Then $f(x)$ is   $\bf{My\; Try::}$ Using $$f'(x) = \lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h} = \lim_{h\rightarrow 0}\frac{f(x)+f(h)+h\sqrt{f(x)}-f(x)}{h}$$  Now Put $x=y=0$ in  $$f(x+y)=f(x)+f(y)+y\sqrt{f(x)}\;,$$ We get $f(0)=0$  So we get $f(0)=0$  So $$f'(x) = \sqrt{f(x)}+\lim_{h\rightarrow 0}\frac{f(h)}{h}=\sqrt{f(x)}$$  So $$\int\frac{f'(x)}{\sqrt{f(x)}}dx = 1\int dx\Rightarrow 2\sqrt{f(x)}=x+c$$  Now Put $x=\;,$ We get $c=0$  So we get $2\sqrt{f(x)}=x\Rightarrow 4f(x)=x^2\Rightarrow \displaystyle f(x)=\frac{x^2}{4}$  Can we solve it some short way, If yes then please explain here, Thanks
2883
Problem:  $$y''+20y=100\sin(wt)$$  conditions:  $$y(0)=2, y'(0)=0$$  What would the solution for $y(t)$ be when $w=2\sqrt 5$  So far I have GS(H)=Acos($\sqrt 20$)+Bsin($\sqrt 20$).
110862
Let $f$ be a differentiable function satisfying the functional time $f(xy)=f(x) +f(y) +\frac{x+y-1}{xy} \forall x,y \gt 0 $ and $f'(1)=2$  My work   Putting $y=1$  $$f(1)=-1$$ $$f'(x)=\lim_{h\to 0}\frac{f(x+h) - f(x)}{h}$$ But I don't know anything about $f(x+h)$  so what to do in this problem ?
31978
$$y' = 2 - \sin(xy), \qquad\quad 1 \leq x \leq 3, \qquad\quad y(1) = -\frac{1}{2}$$  Attempt:=  $$|y''(x)| = |-\cos(xy)(y + xy')| \leq |y + xy'|$$  Not sure what to do next.
55876
Consider equations defined in $D_x = (0,\infty) \times (0,\infty)$  1) Find the maximal solution to  $$ \dot y = \frac{3t^2}{5y^4} $$ satisfying the initial value problem $y(1) = 2^{\frac{1}{5}}$  I have found the maximum solution to be $((0,\infty), (t^3+1)^\frac{1}{5})$  2) Find the maximal solution to   $$ \dot y = -\frac{2y}{5t}+\frac{3}{5y^4}$$  satisfying the IVP $y(1) = 2^{\frac{1}{5}}$.  (HINT: Use a transformation of the form $t^{\alpha}y(t)$.   I hope someone can explain how to use the transformation to find the maximal solution.
6910
It is asked to solve the PDE  $$u_x + u_y + u = e^{x+2y}$$  My attempt: We have that $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u} $$  $\Rightarrow \frac{dy}{dx}=1 \Rightarrow y=x+C_1 \iff C_1=y-x$  $\Rightarrow \frac{du}{dx}= e^{x+2y}-u \iff \dot{u}+u=e^{x+2y} \iff \\ \frac{d}{dx}(e^x u) = e^{2x+2y} \iff u = e^{-x} ( \frac{1}{2}e^{2x+2y} + C_2)$  As $u$ is constant along its characteristics:  $$u(x,y)=\frac{e^{-x}}{2}(e^{2x+2y} + 2C_2(y-x))$$  where $C_2$ is any function of one variable. But wolfram gave the following answer  $$u(x,y)=\frac{e^{-x}}{4}(e^{2x+2y} + 4C_2(y-x))$$  as you can check   What am I doing wrong?
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?
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.