Solving differential equation y "+ 9y = sin3x Thank you very much. Let's go back quickly. The answer is y = c1sin3x + c2sin3x-1 / 6xsin3x

RELATED INFORMATIONS

• 1. Y "+ 9y = xsin3x to solve differential equations
• 2. Find the special solution of the differential equation 4Y "- 12Y '+ 9y = 0 satisfying the initial conditions y ▏ y = 0 = 1, y' ▏ x = 0 = 1 The comma above is an apostrophe
• 3. There are two kinds of differential equations in the reducible higher order differential equations: y '' = f (x, y ') and y' '= f (y, y') The former equation can be solved by Y '= P, then y' = DP / DX = P ', while the latter equation can be solved by Y' = P, then y '= P * DP / dy. for example: YY' - y '^ 2 = 0, this equation is expressed by Y' = P, y '= P * DP / dy. why is y' = 1 + y '^ 2 solved by Y' = P, y '= P'? I know that, but the examples I gave are all short of X-type. Why are they different?
• 4. The reducible higher order differential equation XY '+ 1 = y' ^ 2
• 5. The second derivative of y to x equals ax & sup3; + B I know that first let the first derivative of y be equal to u, and then change it into the first derivative of u by substitution, but then a polynomial with the square of u equal to x appeared, and I can't do it any more Seek expert solution Wrong, it's equal to ay & sup3; + B Otherwise it would be too simple~
• 6. On the problem of solving second order non homogeneous linear differential equation with constant coefficient in postgraduate entrance examination mathematics It is known that the equation is a second-order non-homogeneous linear differential equation with constant coefficients, and it has two special solutions: Y1 = cos2x-1 / 4xsin2x, y2 = sin2x-1 / 4xsin2x. Now the expression of the equation is required. Let the general solution of the equation be y = C1 * cos2x + C2 * sin2x - 1 / 4sin2x, where C1 and C2 are arbitrary constants, Cos2x and sin2x should be a special solution of the corresponding homogeneous differential equation, - 1 / 4sin2x should be a special solution of this equation, but the problem does not give these two conditions, ah, hope to have a great God to help solve, thank you!
• 7. A mathematical differential equation problem (x + 1) y '+ 1 = 2 (to the - y power of E) Ask for his general understanding
• 8. Does the content of the third postgraduate entrance examination of mathematics include the higher order differential equation that can reduce the price
• 9. For the point (x0, Y0, Z0), t tends to 0, and the function f () satisfies f（x0+t,y,z)=f(x0,y0,z0)*P(y-y0,z-z0）； Where p () is a two-dimensional normal distribution function related to y-y0 and z-z0, Given the initial value of F (x0, Y0, Z0), I want to find the value of any f (x1, y, z) at x = x1, as long as I think about it If you can do it, I'll give you my share The fairy on the first floor f(x0,y0,z0)*P(y-y0,z-z0）-f（x0,y,z) =F (x0, Y0, Z0) (P (y-y0, z-z0) - 1) is not very right, I'm changing f (x0 + T, y, z) = f (x0, Y0, Z0) * P (y-y0, z-z0) × exp (- at); a is a known constant
• 10. Solving differential equation y '= cosx / Y
• 11. Find the solution of the differential equation y ^ (4) + y '' '+ y' + y = 0,
• 12. The general solution of the differential equation y "+ 8y '+ 16y = 0 is?
• 13. Find the general solution of the differential equation y '' '+ 8y = 0
• 14. Solving differential equation y * y '' - (y ') ^ 2-y ^ 2 * y' = 0
• 15. Special solution of differential equation y '= y satisfying initial condition y (0) = 1 Right away
• 16. Find the solution of the differential equation y '' = y'e ^ y satisfying the condition y (0) = 0, y '(0) = 1
• 17. Special solution of differential equation y '+ y = 1 satisfying initial condition y (0) = 0
• 18. Finding the general solution of differential equation y '' + 2Y '= - x + 3
• 19. General solution of differential equation y "+ 2Y = x
• 20. General solution of differential equation y '' - 2Y '= x