Question
Apply the method of variation of parameter to solve the differential equation:
Answer :
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The given differential equation is: [ y'' + 6y' + 9y = \frac{1}{x^3} e^{-3x}, \quad x>0 ] The associated homogeneous equation is: [ y_h'': y'' + 6y' + 9y = 0 ] The characteristic equation is: [ r^2 + 6r + 9 = 0 ] [ (r + 3)^2 = 0 \implies r = -3 \text{ (repeated root)} ] So the complementary solution is: [ y_c = (C_1 + C_2 x)e^{-3x} ] For variation of parameters, let the particular solution be: [ y_p = u_1(x) y_1 + u_2(x) y_2 ] where (y_1 = e^{-3x}) and (y_2 = x e^{-3x}). The formulas are: [ u_1' = - \frac{y_2 g(x)}{W}, \quad u_2' = \frac{y_1 g(x)}{W} ] with (g(x)
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Related Questions
Using the method of variation of parameters, solve the equation
Using Charpit's method, solve the equation
हल कीजिए :
in
is a linear homogeneous equation
प्राचल विचरण विधि द्वारा समीकरण$
का हल प्राप्त कीजिए।
Using the method of undetermined coefficients, solve the differential equation:
The Pfaffian equation is integrable.