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

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› 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)

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› 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)

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› 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)

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› 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

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› 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

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= −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