I was dizzy when I did the question. Ask the bigwigs for help. Ordinary differential equation proble

Yes, the integration of both sides, this is the most basic thing, it is equivalent to the derivation on both sides of the equation on the right at the same time becomes the formula on the left, it may be that the time to do the problem is too long, the brain is stuck, just take a break.

Solution: The differential equation is dy dx= (1-y), which is reduced to.

dy/√(1-y²)=dx

There is arcsiny=x+c (c is an arbitrary constant), y=sin(x+c) x0)=y0

The general solution of the arcsiny0=x0+c,c=arcsiny0-x0 equation is.

y=sin(x+arcsiny0-x0)

Refer to the chapter on Differential Equations in Advanced Mathematics. If not, I'll send it to you. To put it simply, the solution of ordinary differential equations is to find the eigenroot such as y''-y'-2y =0 Its eigenequation corresponds to r 2 - r -2 0 (this can be written, and the above corresponds) The eigenroot is r= 2 , -1 The next step is to write the general solution y= c1*e (2x) +c2*e (-x) according to the eigenroot

For the case of double roots and complex roots, the general solution is relatively complex, please refer to "Advanced Mathematics" If you know the two boundary conditions, you can find the value of c1 c2. The above is the solution of a homogeneous ordinary differential equation, and this solution is called a homogeneous solution. For non-homogeneous, its solution y = homogeneous solution + special solution as y''-y'-2y = e x we can set the special solution according to the format of e x as y*=a*e x (look at the book for the specific format) into the above formula, we can know that a= can be known to be solved as y = c1*e (2x) +c2*e (-x).

If you integrate directly, 2xdp+pdx cannot be written as a full differentiation, and it cannot be written.

And p*(2xdp+pdx) = x d(p 2) +p 2) dx = d(x*p 2) can be written as full differential.

This is a non-homogeneous differential equation that requires two linearly independent solutions to its corresponding homogeneous differential equation:

Y3-Y1 and Y2-Y1

So the general solution of the homogeneous differential equation is:

The general solution of the non-homogeneous differential equation of C1(Y3-Y1) +C2(Y2-Y1) = the general solution of the homogeneous differential equation + the special solution of the non-homogeneous differential equation is :

c1(y3-y1) +c2(y2-y1) +y1 is substituted into the above equation and is generally explained as:

y = (c1 + c2x)e^2x + x

xy''-y'=0，y''/y'=1/x

Points to get LNY'=LNX+LNC1, so y'=c1x。

Continue to integral, y=c1 2 x 2+c2.

If you can think of c1 2 as the new c1, then the general solution can also be written as y=c1 x 2+c2.

I really want to do it, and I feel familiar.

<> "I hope there will be a roll, and the jujube will be demolished to help tease."