? asked in Science & MathematicsEngineering · 7 years ago

Second order differential equation using D-Operators?

d^2y/dt^2+2 dy/dx+7y=e^-tcos2t


Plzzz help me to solve this equation.I have a problem in finding the particular integral if g(t) is a product

2 Answers

  • 7 years ago
    Favorite Answer

    I assume you meant to write :

    d²y/dt² + 2.dy/dt + 7y = ℮^(-t).cos(2.t) i.e. no variable x in the equation.

    Considering the solution of the corresponding homogeneous equation. the auxiliary equation is :

    m²+2m+7 = 0 or m₁= -1+j√6 and m₂=-1-j√6

    which gives the general solution :

    yh = ℮^(-t).[A.cos(√6.t) + B.sin(√6.t)]

    I should check this back to see that it satisfies the differential equation (LHS) but no time I'm afraid!

    We now have to find any particular solution yp to add to this in order to get a general solution of the non-homogeneous equation. Since the DE has constant coefficients, the method of undetermined coefficients should do this for us!

    Try yp = ℮^(-t)[K.cos(2.t) + M.sin(2.t)]

    yp' = ℮^(-t)[-2.K.sin(2.t) + 2.M.cos(2.t) – K.cos(2.t) – M.sin(2.t)] and

    yp" = ℮^(-t)[-4.K.cos(2.t) – 4.M.sin(2t) + 2.K.sin(2t) – 2.M.cos(2.t) + 2.K.sin(2.t) - 2.M.cos(2,t) + K.cos(2.t) + M.sin(2.t)]

    Subtitute these in the DE and equate coefficients of cos(2.t) and sin(2.t) to obtain values for K and M (2 equations, 2 unknowns!). You should reach a particular solution which can simply be added to the above general solution of the homogeneous equation to give the complete solution y = yh + yp

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  • Anonymous
    4 years ago

    a) i) First sparkling up m^2 + 5m + 6 = 0 Giving suggestions of m= -2 and m= -3. the final answer is for this reason y = Ae^(-2x) + Be^(-3x) ii) sparkling up m^2 + 8m +sixteen = 0 Giving answer of m= -4 the final answer is y = (A + Bx)e^(-4x) iii) sparkling up m^2 + 4m + 8 = 0 (m+2)^2 +4 = 0 m+2 = +/-sqrt(-4) m+2 = +/-sqrt4*sqrt(-a million) m+2 = +/-2i m = -2 +2i or m= -2 - 2i So typical answer is y = e^(-2x) (Acos2x + Bsin2x) those are the three possible varieties of the suggestions depending on no count if the quadratic in m has a million, 2 or 0 genuine suggestions. purely plug interior the suggestions to this equation interior the situation I even have and you additionally can sparkling up the different equation of this type. b) The P.I. would be of the style y=acos(2x)+bsin(2x). this provides: dy/dx = -2asin(2x) + 2bcos(2x) and: d^2y/dx^2 = -4acos(2x) -4bsin(2x) Plug this into the unique equation: -4acos(2x) - 4bsin(2x) + 4(-2asin(2x) + 2bcos(2x)) + 8(acos(2x) + bsin(2x)) = 4sin(2x) - 12cos(2x) Now study coefficients of sin(2x): -4b - 8a + 8b= 4 4b -8a = 4 Now study the coeffs of cos(2x): -4a + 8b + 8a = -12 4a + 8b = -12 Now sparkling up those concurrently to grant a=-1and b=-a million So the P.I. is -cos(2x) -sin(2x) the final answer is y=c.f + P.I. The c.f. is comparable to we got here across partly a.iii. So the final answer is: y = -cos(2x) - sin(2x) + e^(-2x) (Acos(2x) + Bsin(2x)) c) while x=0, y=a million a million = -cos0 - sin 0 + (Acos0 +Bsin0) a million = -a million + A A = 2 while x=0 y'=-6 y' = 2sin2x - 2cos2x +e^(-2x)(-4sin2x+2Bcos2x) - 2e^(-2x)(2cos2x+Bsin2x) -6 = -2 -4 +2B B=0 So specific answer is: y = -cos(2x) - sin(2x) + e^(-2x) (2cos(2x))

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