The solution y =? Of the differential equation y '' + 2Y '+ y = 0?

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• 1. Finding the special solution of differential equation y '- 2Y / (1-x ^ 2) = x + 1, x = 0, y = 0
• 2. Find the special solution of y = y '= 0 when y' '+ (y)' ^ 2 = 1 and x = 0
• 3. 1. A straight line is cut by two straight lines L1: 4x + y + 6 = 0, L2: 3x-5y-6 = 0, and the midpoint of the line segment is P point. When the coordinate of P point is (0,0), find the straight line 1. A straight line is cut by two straight lines L1: 4x + y + 6 = 0, L2: 3x-5y-6 = 0, and the midpoint of the line segment is P point. When the coordinate of P point is (0,0), the linear equation is solved 2. If the line L and two lines y = 1, x-y-7 = 0 intersect at P and Q respectively, the middle point of the line PQ is (1, - 1) the equation for solving L
• 4. The tangent equation of curve y = x ^ 3 + 3x-8 at x = 2 is We need to explain it in detail
• 5. Find the tangent and normal plane equation of the curve X ^ 2-y ^ 2 = 3 and x ^ 2 + y ^ 2-z ^ 2 = 4 at the point (- 2, - 1,1)
• 6. Finding tangent and normal plane equation of curve y ^ 2 = 2mx Z ^ 2 = M-X at point (x.y.z.)
• 7. Finding the tangent equation of curve X ^ 2 + y ^ 2 + Z ^ 2 = 2, x + y + Z = 0 at point (1,0. - 1) is a normal plane equation
• 8. Find a point on the curve Γ: x = t, y = T ^ 2, z = T ^ 3 so that the tangent of the point is parallel to the plane π: x + 2Y + Z = 4
• 9. Find the tangent equation and normal plane equation of the curve X = t, y = T ^ 2, z = T ^ 3 at t = 2
• 10. Among all the tangents of the curve X = t, y = - T ^ 2, z = T ^ 3, how many are parallel to the plane x + 2Y + Z = 4? Hello teacher, my solution is like this: 1 find the tangent vector (1, - 2T, 3T ^ 2) 2 orthogonal to the normal vector of the plane (1,2,1) can get 1-4t + 3T ^ 2 = 0, the solution is t = 1 / 3, t = 1, so there are two, but what I don't understand is why the answer adds a restriction, that is, after finding t, bring in the curve to get the tangent point P1 (1 / 3, - 1 / 9, 1 / 27), P2 (1, - 1, 1) Because the tangent point is not on the plane, the final answer is two. What I want to ask is that even if the tangent point is on the plane, the tangent vector within the plane is parallel to the plane. Why should it be regarded as a limiting condition
• 11. Finding the special solution of differential equation y '- y = cosx, x = 0, y = 0
• 12. Differential equation solution: y '- ysinx - y ^ 2 + cosx = 0
• 13. Solve the differential equation (x ^ 2-1) y '+ 2XY cosx = 0, thank you very much! (x ^ 2-1) y & # 39; + 2XY cosx = 0, the garbled part is the derivative of Y, detailed process
• 14. Which differential equation is y = cosx
• 15. Solving differential equation y '= cosx / Y
• 16. 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
• 17. Does the content of the third postgraduate entrance examination of mathematics include the higher order differential equation that can reduce the price
• 18. A mathematical differential equation problem (x + 1) y '+ 1 = 2 (to the - y power of E) Ask for his general understanding
• 19. 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!
• 20. 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~