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Method of Variation of ParametersBy Kaushal Patel
Method of Variation of Parameters
› Langrage invented the method of variation of parameters.
› Consider differential equation of the form f(D)y=X.
› when X is of the form 𝑒𝑎𝑥 , sin 𝑎𝑥, cos 𝑎𝑥, 𝑥𝑚, 𝑒𝑎𝑥 . 𝑣 or any function of x, then the shortcut methods are available which will discuss later on. If X be of any other form say tan 𝑥, sec 𝑥 , csc 𝑥 𝑒𝑡𝑐. , then we have to use one of the following methods.
I. The method of partial fractions
II. The method of variation of parameters
Continue…
› Consider the second order linear differential equation with constant co-efficient
𝑑2𝑦
𝑑𝑥2+ 𝑎1
𝑑𝑦
𝑑𝑥+ 𝑎2𝑦 = 𝑋 ……(i)
› Let the complementary function of (i) be 𝑦 = 𝑐1𝑦1 + 𝑐2𝑦2
› Then 𝑦1 and 𝑦2 satisfy the equation
𝑑2𝑦
𝑑𝑥2+ 𝑎1
𝑑𝑦
𝑑𝑥+ 𝑎2𝑦 = 0 ……(ii)
Continue…
› Let us assume that the particular integral of (i) be
𝑦 = 𝑢𝑦1 + 𝑣𝑦2 ……(iii)
where u and v are unknown functions of x
› Differentiating w.r. to x, we have
𝑦′ = 𝑢𝑦1′ + 𝑢′𝑦1 + 𝑣𝑦2
′ + 𝑣′𝑦2 ……(iv)
› To determine two unknown functions u and v, we need two equations.
We assume that 𝑢′𝑦1 + 𝑣′𝑦2 = 0 ……(v)
∴(iv) reduces to
𝑦′ = 𝑢𝑦1′ + 𝑣𝑦2
′ ……(vi)
Continue…
› Differentiating w.r. to x, we get𝑦" = 𝑢𝑦1
" + 𝑢′𝑦1′ + 𝑣𝑦2
" + 𝑣′𝑦2′
› Substituting the values of 𝑦, 𝑦′ and 𝑦" in (i), we have
𝑢 𝑦1" + 𝑎1𝑦1
′ + 𝑎2𝑦1 + 𝑣 𝑦2" + 𝑎1𝑦2
′ + 𝑎2𝑦2 + 𝑢′𝑦1′ + 𝑣′𝑦2
′ = 𝑋 ……(vii)
› But 𝑦1 and 𝑦2 satisfy equation (ii)
∴ 𝑦1" + 𝑎1𝑦1
′ + 𝑎2𝑦1 = 0 and 𝑦2" + 𝑎1𝑦2
′ + 𝑎2𝑦2 = 0
› Equation (viii) takes the form,
𝑢′𝑦1′ + 𝑣′𝑦2
′ = 𝑋 ……(viii)
Continue…
› Solving (v) and (ix), we get
𝑢′ =𝑦2𝑋
𝑦1𝑦2′−𝑦2𝑦1
′ and 𝑣′ =𝑦1𝑋
𝑦1𝑦2′−𝑦2𝑦1
′
› Integrating, we get
𝑢 = − 𝑦2𝑋
𝑤𝑑𝑥 and v = −
𝑦1𝑋
𝑤𝑑𝑥, where 𝑤 = 𝑦1𝑦2
′ − 𝑦2𝑦1′
› Substituting the values of u and v in (iii), we have
𝑃. 𝐼 = 𝑦 = −𝑦1 𝑦2𝑋
𝑤𝑑𝑥+𝑦2
𝑦1𝑋
𝑤𝑑𝑥 ……(ix)
Example
Find general solution of 𝑦" + 9𝑦 = sec 3𝑥 by method of variation of parameters.
Solution:-
› The given differential equation can be written as,
𝐷2 + 9 𝑦 = sec 3𝑥 , 𝑤ℎ𝑒𝑟𝑒 𝐷 =𝑑
𝑑𝑥
› The auxiliary equation is,𝐷2 + 9 = 0
=>𝐷 = ±3𝑖
Continue…
› Hence, complimentary function is𝑦𝑐 = 𝑐1 cos 3𝑥 + 𝑐2 sin 3𝑥
› Now we take 𝑦1 = cos 3𝑥 , 𝑦2 = sin 3𝑥
𝑤 =𝑦1 𝑦2𝑦1′ 𝑦2
′ =cos 3𝑥 sin 3𝑥−3 sin 3𝑥 3 cos 3𝑥
=3
› Particular integral is,
𝑦𝑝 = −𝑦1 𝑦2𝑋
𝑤𝑑𝑥 + 𝑦2
𝑦1𝑋
𝑤𝑑𝑥
Continue…
= −cos 3𝑥 sin 3𝑥 sec 3𝑥
3𝑑𝑥 + sin 3𝑥
cos 3𝑥 sec 3𝑥
3𝑑𝑥
= −cos 3𝑥
3 tan3𝑥 𝑑𝑥 +
sin 3𝑥
3 𝑑𝑥
=cos 3𝑥 log(cos 3𝑥)
9+𝑥 sin 3𝑥
3
› Hence, general solution is
y(x)= 𝑦𝑐 + 𝑦𝑝
= 𝑐1 cos 3𝑥 + 𝑐2 sin 3𝑥 +cos 3𝑥 log(cos 3𝑥)
9+𝑥 sin 3𝑥
3