y 3y 2sinx special solution or general solution

Homogeneous eigenequations.

r^2+3=0

r=±√3i

The homogeneous general solution y=c1cos( 3x) + c2sin ( 3x) and the special solution is y=asinx+bcosx

y'=acosx-bsinx

y''=-asinx-bcosx

Substituting the original equation is obtained.

asinx-bcosx+3(asinx+bcosx)=2sinx.

2a=2,b=0

a=1 is y=sinx

The general solution of the original equation is y=c1cos( 3x) + c2sin( 3x) + sinx

Homogeneous equation y''The characteristic equation of +3y=0: r 2 + 3 = 0 r = root number 3i or r = - root number 3i

Therefore, the above general solution: y=c1cos (root number 3x) + c2sin (root number 3x) since 2sinx can be written as: 2sinxe (0*x) where the x coefficient of sinx is 1, e (0*x)x coefficient 0, 0+i is not the root of the characteristic equation.

Therefore, the special solution is assumed: y*=(acosx+bsinx)y'=-asinx+bcosx

y''=-acosx-bsinx

Substituting the original equation:

acosx-bsinx)+3(acosx+bsinx)=2sinx2acosx)+2bsinx=2sinx

a=0 b=1

y*=sinx

General solution: y=c1cos (root number 3x) + c2sin (root number 3x) + sinx

Homogeneous leakage is like a slip equation y''+3y=0 characteristic arubber loss equation: r 2+3=0 r = root number 3i or r = - root number 3i so the above pass y=c1cos (root number 3x) + c2sin (root number 3x) since 2sinx can be written as: return wax 2sinxe (0*x) where the x coefficient of sinx is 1, e (0*x)x coefficient 0, 0+i is not the root of the characteristic equation.

Therefore, the special y*=(aco....)

Summary. Kiss <>

Hello, happy to answer your <>

The general solution of y = e +5x is: y(x)=y h(x)+y p(x)=ax+b+e +x 2+6x. Find the corresponding homogeneous equation y first''=0.

Since its eigenequation is r 2=0$, its general solution is y h(x)=ax+b, where a, b are arbitrary constants.

y = e +5x.

Good drops. Kiss <>

Hello, I'm glad to answer for you the general solution of [Fresh Sensitive Flowers]y =e +5x is: y(x)=y h(x)+y p(x)=ax+b+e +x 2+6x. Find the corresponding homogeneous equation y first''=0.

Since its characteristic square bridge hail auspicious distance is r 2=0$, its general solution is y h(x)=ax+b, where a, b are arbitrary constants.

Kiss <>

Okay. Can you tell us about the process.

The whole process. Kiss <>

Expansion: Non-homogeneous group Lucheng hail zone $y''=e +5x 2$. Bring $y p(x)$ into the equation to get y''p=2c+0+0=2c e +5x 2=c(2x) 2+dx+e Substituting x=0 into the above equation gives c=e 0=1.

Take the first derivative on both sides of the equation at the same time to get 6e +10x=4cx+d, so that x=0 gives d=6. Again, taking the second derivative on both sides of the equation at the same time gives $18e +10=4c, so c=+10}.

Kiss [Big Red Flowers]: Non-Qi Bird Quarrel Stupid Equation Y''A special solution of =e +5x 2 is: y p(x)=e + fracx 2+6x the original equation y''The pass sedan of =e +5x 2 is explained as:

y(x)=y h(x)+y p(x)=ax+b+ frace + fracx 2+6x where a, b are arbitrary constants.

Kiss, you can take a look at the process.

Kiss, the teacher answers the detailed process.

Summary. Pro, y=2x +c x+c

y''+y=4x solution.

Pro, y=2x +c x+c

Is there a process. Dear, my solution steps:1

Convert the equation to a standard form: y''+y=4x³2.Let the special solution be in the form y=x(x), and bring in the original equation to get the differential equation of x(x).

x''(x)=4x³3.The solution is x(x)=2x +c x+c 4Since the original equation has two unknowns c and c, the special solution is y=2x +c x+c

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y = 2x.

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y=1/2x⁴+c

Is there a process. y = 2x because y = 2x so dy=2x dx dy= 2x dxy=1 2x +c

y = 2x because y = 2x so dy=2x dx dy= 2x dxy=1 2x +c

What about the general explanation of this?

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Firstly, the equation is reduced to the form of the slag resistance separation variable, and the shift term is obtained:

e^(2x)dy = y+1)dx

Then divide both sides by (y+1)e (2x) at the same time to get :

1 (y+1) dy = 1 e (2x) dx integrates both sides at the same time, yielding:

ln|y+1|=1 2 e (-2x) +c, where c is constant. Simplify the above formula to obtain:

y = 1 + ce^(-2x)

This is the general understanding of the original army's attack.

Solution: Homogeneous equation y''-8y'The characteristic equation of +16y=0 is r -8r+16=0, then r=4

Homogeneous equation y''-8y'The general solution of +16y=0 is y=(c1x+c2)e (4x).

c1, c2 integral constants).

Let the solution of the original equation be y=ax e (4x).

y'=4ax²e^(4x)+2axe^(4x)

y''=16ax²e^(4x)+16axe^(4x)+2ae^(4x)

Substituting the original equation yields 2ae (4x) = e(4x).

>2a=1

>a=1/2

A special solution of the original equation is y=x e (4x) 2

The general solution of the original equation is y=(c1x+c2)e (4x)+x e (4x) 2

c1, c2 integral constants).

The initial condition is y(0)=0,y'(0)=1

Substituting the general solution yields c1=1 and c2=0

Therefore, the solution of the original equation satisfying the initial condition is y=(x+x 2)e (4x).

Note: The initial condition y(0)=0,y'(0)=1 is used to determine the integral constants c1 and c2.

Obviously, y=0 is the solution of the original equation.

If y≠0, (x-2xy-y 2)y'+y^2=0

>y^2dx/dy+(1-2y)x=y^2.(1) Equation (1) is about y-order linear differentiation.

Therefore, from the general solution of the first-order linear differential equation, the general solution of equation (1) is x=y 2(1+ce (1 y)).

c is a constant), so the general solution of the original equation is y=0 and x=y2(1+ce (1 y)).