Find the solution of the differential equation y 3y 2y e to the xth power

Hello landlord, this is a non-homogeneous second-order linear differential equation.

The solution is divided into two parts, one is the general solution y0 obtained by applying the formula so that the right side is equal to 0, and the other part is the special solution y1 of the given equation. The specific solution is as follows:

The characteristic equation is +2 -3=0 and the solution is =1,-3.

So the general solution of the equation is y0=ae x+be (-3x) assuming the special solution y1= e 2x, brought into the original equation, 4 e 2x+4 e 2x-3 e 2x=e 2x

5αe^2x=e^2x,α=1/5

So the general solution of the original equation y=y0+y1=1 5e 2x+ae x+be (-3x) (a, b are arbitrary constants).

Solution: Homogeneous equation y''+3y'The characteristic equation for +2y=0 is r 2+3r+2=0, then.

r1=-1，r=-2

The general solution of this homogeneous equation is y=c1e (-x)+c2e (-2x)c1, c2 is constant)).

Let the solution of the original equation be y=ae x and substitute it into the original equation.

ae^x+3ae^x+2ae^x=e^x

a=1 6y=e x 6 is a special solution of the original equation.

Therefore, the general solution of the original equation is y=c1e (-x)+c2e (-2x)+e x 6.

y''－3y' +2y = e^(3x)

The eigenequation r 2 - 3r + 2 = 0, the eigenroot r = 1, 2

Therefore, if the special solution y = ae (3x), substituting the differential equation into the school of 2a = 1, a = 1 2, then the general undisturbed solution of the differential equation is y = c1e x + c2e (2x) +1 2)e (3x).

Hello friends! The detailed process rt, I hope that the repentance can help you solve the problem of the banquet.

Summary. Hello, I am glad to answer for you, find the differential equation y,,4y = sin 2x + e Because the non-homogeneous term is sin2x, 0 2i is the eigenroot, so a special solution of the original equation can be set as y*=x(acos2x+bsin2x), consultation record · on 2022-08-29

Find the general solution of the differential equation y,,4y = sin 2x + e.

Hello, I am glad to solve for you, find the differential equation y ,,4y = sin 2x + e because the non-homogeneous term is sin2x, 0 2i is the eigenroot, so a special solution of the original equation can be set as y*=x(acos2x+bsin2x), y*'=acos2x+bsin2x)+x(2bcos2x-2asin2x)=(a+2bx)cos2x+(b-2ax)sin2x，y*''2BCOS2X-2(A+2BX)Sin2X-2ASIN2X+2(B-2AX)cos2x=4(B-AX)CoS2X-4(A+BX)sin2X, simplification yields 4BCOS2X-4ASIN2X=Sin2X, so 4B=0,-4A=1, i.e., A=-1 4,B=0, so a special solution of the original equation is y*=(1 4)xcos2x

I want to figure it out. Find the differential equation y''4y = general solution of sin 2x + e.

Hello, I am glad to answer for you, first find the homogeneous linearity, the eigenroot equation r +4=0 to get the general solution y=c(x)(c1sin2x+c2cos2x), from c (x)(c1sin2x+c2cos2x)=sin2x to find c(x) to get the general solution y*=(1 4)xcos2x

Summary. Since the right side of a nonhomogeneous equation is a polynomial plus an exponential function, we can guess that its special solution is a function of the shape ax + be x. Substituting the original equation gives a = 1 2 and b = 1 2, so the special solution is -x 2 + e x 2.

In the end, the general solution for u is u = ce (-cos(x)) x 2 + e x 2, and the general solution for y is y = ce (-cos(x)) x 2 + e x 2) cos(x).

Differential equation y'sin+y=2x+e.

Hello dear! This is a first-order non-homogeneous linear differential equation that can be solved using the constant variation method.

First, simplify the equation: Replace y'sin shifts get y'=2x + e -ysin) cos, and then let the wild dig u = y*cos, then the original equation can be expressed as u' +usin = 2x + eˣcos。Next, we need to find the general solution of u.

For homogeneous equation u'+usin = 0, the identity u = ce (-cos(x)) can be used to find its general solution (where c is an arbitrary constant that destroys only).

Since the right side of the non-homogeneous equation is a polynomial repentance plus an exponential function, I can guess that its special solution is a function of the form ax + be x. Substituting the original equation gives a = 1 2 and b = 1 2, so the special solution is -x 2 + e x 2. In the end, the general solution for u is u = ce (-cos(x)) x 2 + e x 2, while the pre-scrambling solution for y is y = ce (-cos(x)) x 2 + e x 2) cos(x).

Original equation y''-3y'The homogeneous equation corresponding to +2y=x(e x) is y''-3y'+2y=0

The characteristic equation is that the land bend is r -3r+2=0

The eigenroots are r(1)=1 and r(2)=2

So the solution of the homogeneous equation is y(1)=c(1)(e x)+c(2)[e (2x)];

Since 1 is the eigenroot, let the special solution of the original equation be y(2)=(ax +bx)(e x).

then y'(2)=(2ax+b)(e^x)+(ax²+bx)(e^x)

ax²+(2a+b)x+b](e^x)

y"(2)=[2ax+(2a+b)](e^x)+[ax²+(2a+b)x+b](e^x)

ax²+(4a+b)x+(2a+2b)](e^x)

Substituting the original equation, then.

ax²+(4a+b)x+(2a+2b)](e^x)-3[ax²+(2a+b)x+b](e^x)+2(ax²+bx)(e^x)=x(e^x)

i.e. (4a+b)-3(2a+b)+2b=1

2a+2b)-3b=0

The solution gives a=-1 2, b=-1

That is, the special solution is y(2)=[1 2)x -x](e x).

Thus, the general solution of the original equation is.

y=y(1)+y(2)

c(1)(e^x)+c(2)[e^(2x)]+1/2)x²-x](e^x)

(1/2)x²-x+c(1)](e^x)+c(2)[e^(2x)]

Not verified, please be cautious, 1, blue126 report.

Since 1 is the eigenroot, let the special solution of the original equation be y(2)=(ax +bx)(e x) This early step is to dig and not understand this ......Why do you want to make it ax + bx?

Report 4768858

I can't remember the reason now, it's said in the textbook, you can look at it Okay, thank you......,

The characteristic equation of the second-order differential equation y +3y +2y=0 is: r2-2r-3=0, and its eigenroot is: r1=3, r2=-1, because e-x =-1, is the single root of the corresponding eigenequation, and it can be known from the properties of the differential equation:

The form of the special solution is: axe-xSubstituting the special solution into the original equation obtains: -2ae-x+axe-x+ae-x-axe-x+2a....

dy dx-2y=e x(1) first find the general solution of the homogeneous differential equation dy dx-2y=0. The characteristic root of the equation satisfies -2=0, and =2 is obtained, so the Qi mountain or the subdifferential equation is solved by y=ce (2x)(2), and then the special kernel of the non-homogeneous differential equation is disassembled. Define the differential operation d dx=d, 1 d= , where l(d)=d-2, then the special solution y* has (d-2)y*=e x

Hence y*=.

Let the special solution be y2=ax 2e (3x).

then y2'=(3ax^2+2ax)e^(3x)y2''=9ax 2+12ax+2a)e (3x)So 9ax 2-18ax 2+9ax 2+12ax-12ax+2a=1

So a=1 2

y2=x^2e^(3x)/2

To find the general solution of the differential equation y -2y -3y=e x: first solve the corresponding homogeneous differential equation y -2y -3y=0 and its characteristic equation is: r 2-2r-3=0

The eigenroots are: r 1 = 3 and r 2 = -1

So the general interpretation is: y=c 1e +c 2e

Then solve a special solution of the differential equation y -2y -3y=e x, let its special solution be y=ae x, and substitute the equation to solve a=-4, so the general solution of the differential equation y -2y -3y=e x is: y=c 1e +c 2e -4e x