General solution of differential equation y '' + y = x + cosx

General solution of differential equation y '' + y = x + cosx

The general solution of Y '' + y = 0 is y = c1sinx + c2cosx, the special solution of Y '' + y = x is y = x, and the special solution of Y '+ y = cosx is y = X
Y = x (acosx + bsinx), then y '= acosx + bsinx + X (bcosx asinx),
Y '' = 2bcosx-2asinx-x (bsinx + acosx)
2bcosx-2asinx = cosx, B = 0.5, a = 0, the special solution is y = 0.5xsinx
In conclusion, the general solution is y = C 1 SiN x + C 2 cos x + X (0.5 SiN x + 1)

General solution of differential equation y '+ ysinx = e ^ cosx

∵ homogeneous equation y '+ ysinx = 0 = = > y' = - ysinx
==>dy/y=-sinxdx
==>LNY = cosx + LNC (C is an integral constant)
==>y=Ce^cosx
The homogeneous equation is y = CE ^ cosx (C is an integral constant)
Therefore, let the general solution of the original differential equation be y = C (x) e ^ cosx (C (x) denotes a function of x)
∵y'=C'(x)e^cosx-C(x)sinxe^cosx
Substituting into the original equation, C '(x) e ^ cosx = e ^ cosx
==>C'(x)=1
==>C (x) = x + C (C is an integral constant)
∴y=C(x)e^cosx=(x+C)e^cosx
So the general solution of the original differential equation is y = (x + C) e ^ cosx (C is an integral constant)

General solution of differential equation y '- y = 3x
The general solution of Y '- y = O, y = C1E ^ x is y = u (x) e ^ x, which is reduced to Du / DX = 3x / e ^ X. what's wrong with this solution?

Let y = UE ^ x, y '= u'e ^ x + UE ^ X be substituted by u'e ^ x + UE ^ x = 3Xu' = 3xE ^ (- x) integral: u = ∫ 3xE ^ (- x) DX = - ∫ 3xde ^ (- x) DX = - 3xE ^ (- x) + 3 ∫ e ^ (- x) DX = - 3xE ^ (- x) + 3E ^ (- x)...., if the general solution of Y '- y = O, y = CE ^ X

General solution of Y '= (2x + y) / (3x-y) differential equation

This equation can be transformed into a differential equation with separable variables
Let y = UX
Then dy / DX = XDU / DX + U
Then the original equation can be reduced to
xu'+u=(2+u)/(3-u)
We obtain u '(3-U) / (u ^ - 2U + 2) = 1 / X
The indefinite integral of X on both sides is obtained
∫(3-u)/[(u-1)^2+1]du=lnx+C
We obtain 2arctan (U-1) - 1 / 2ln (u ^ 2-2u + 2) = LNX + C
The general solution is
2arctan(y/x-1)-1/2ln(y²/x²-2y/x+2)=ln+C

How to write the process of 2x-6 / 4-4x + X squared / (x + 3) * x squared + 6x + 9 / 3-x

2x-6 / 4-4x + X squared / (x + 3) * x squared + 6x + 9 / 3-x
=2﹙x-3﹚÷﹙x-2﹚²÷[﹙x+3﹚×﹙x+3﹚²]÷﹙3-x﹚
=2﹙x-3﹚×1/﹙x-2﹚²×1/﹙x+3﹚³×1/﹙3-x﹚
=﹣2／[﹙x-2﹚²×﹙x+3﹚³]

What is the relationship among current, voltage and power?
What is the relationship between power and resistance?

Total power of any circuit (P) = voltage applied at both ends of the circuit (U) * total current in the circuit (I)
This is the definition of electric power, so it must exist for any circuit

Is (1 / x) + (1 / y) a fraction?

The denominator has unknowns
So it's a fraction

In the home circuit, only the 220 V 100 W light bulb is connected, and the light bulb lights up normally, and then the 220 V 800 w refrigerator is connected, and the light bulb turns dark. Why?

Compared with single bulb, when connected to the refrigerator, the current of the power supply circuit increases, the voltage on the power supply circuit increases, the voltage on the electrical appliances (including the bulb) decreases, and all the bulbs darken

What is the common denominator of fractions 3 / A and 2 / 5A? What is the common denominator of fractions 1 / A ^ 2B and 1 / AB ^ 2?

The common denominator of 3 / A and 2 / 5A is 5A
The common denominator of 1 / A ^ 2B and 1 / AB ^ 2 is a ^ 2B ^ 2

The resistance of an electrical appliance is 400 Ω, and the current is 0.6A when it works normally. Now it is connected to a 36V power supply. How to make it work normally

R total = u / I = 36V / 0.6A = 60ohm, because 60ohm ＜ 400ohm, there is another resistor in parallel with it, and the resistance value is x 1 / x + 1 / 400ohm = 1 / 60ohm, x = 70.6ohm, so a 70.6ohm resistor should be connected in parallel