Let s take a look at this problem again, and I still don t understand how to write the expression of

y1 = !(/y0 &&/y2 &&/y4 &&/y6)y2 = !( y1 && y3 && y5 && y7) is as follows.

Input c b a intermediate variable y7 y6 y5 y4 y3 y2 y1 y0 output y1 y2

So y1 = c b a + cb a + c b a + cb a + cb a is simplified to: y1 = a

Y2 = C Ba + CBA + C Ba + CBA Simplified: Y2 = A

y(i)=m(i);i=0~7；Translation of the meaning of i: first convert i to a three-digit binary number, and the zero value of each of its digits is directly determined.

Is it true or false on top'—'such as—

m(i)=a0a1a2 y4=m4=a2 a1 a0

y1 = y0 y2 y4 y6 = Morgan's law = y0 + y2 + y4 + y6 = a2a1a0 + a2a1a0 + a2a1a0 + a2a1a0 + a2a1a0

The same is true for the re-simplification of the Carnot diagram =A0 Non = A Non (note that this a is a in the input a b c).

y2 = y1 y3 y5 y7 = Morgan's law = y1 +y3 +y5 +y7 = a2a1a0 + a2a1a0 + a2a1a0 + a2a1a0 + a2a1a0

The Carnot diagram method is then simplified =a0=a (note that this a is a in the input a b c).

The final result is y1=a, y2=a

Summary. Yes, it is known that ay1+by2+cy3=0 is true if and only if a=b=c=0

Let d(y1-y2)+e(y2-y3)=0 hold, then dy1+(e-d)y2-ey3=0

Because y1y2y3 is linearly independent, d=0, e-d=0, e=0, so d and e are zero. So linear is irrelevant.

y1 and y2 are both functions about x, they are linear and independent, so can the value of y1 be set as a function compared to y2?

Yes, the wild ridge, it can be seen that ay1+by2+cy3=0 is true, if and only if a=b=c=0 lets d(y1-y2)+e(y2-y3)=0 form a stool, then dy1+(e-d)y2-ey3=0 is linear independent of y1y2y3, so d=0, e-d=0, e=0 so d and e are zero. So linear is irrelevant.

So it can be.

So it can be.

OK. What does it have to do with sexual lines?

It was me who made a mistake.

Excuse me. It's okay. You're posting it again.

Because the answer doesn't seem to be the same as what I asked.

Yes, you can.

y''+y'=x^2

y''+y'=0

Characteristic Fang Zheng Fucheng.

r^2+r=0

r=0,r=-1

y=c1e^(-x)+c2

Set y''+y'=x 2 has a special solution y=ax 3+bx 2+cxy'=3ax^2+2bx+c

y''=6ax+2b

6ax+2b+3ax^2+2bx+c=x^23a=1,2b+6a=0 2b+c=0

a=1/3,b=-1,c=2

Special solution y=(1 wide J3) x 3-x 2+2xy''+y'=x 2 solution y=(1 3)x 2-x 2+2x+c1e (-x)+c2

y''+2y'+y=cosx

y''+2y'+y=0

The characteristic equation is called clever with r 2+2r+1=0

r=-1y=c1e (-x)+cxe (-x) set y''+2y'+y=cosx-specific solution y=mcosx+nsinxy'= msinx+ncosx

y''=mcosx -nsinx

mcosx-nsinx-2msinx+2ncosx+mcosx+nsinx=cosx

m+2n+m)=1 (-n-2m+n)=0m=0,n=1/2

y=(1/2)sinx

General solution: y=c1e (-x)+cxe (-x)+(1 2)sinxx=0, y=0 c1=0

y'=-c1e^(-x)+ce^(-x)+cxe^x+(1/2)cosx x=0,y'=-c1+c+(1/2)=3/2 c=1

Solution: y=xe (-x)+(1 2)sinx

Look at the picture. If you have any questions, thank you. If it helps you, you can adopt it haha.

i.e. dy dx = 1+y

There is dy (1+y )=dx

The integral on both sides has arctany = x + c to get y =tan(x+c).

Vedic theorem x1x2 = a part is equal to -a part b