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Question Number 26904 by sorour87 last updated on 31/Dec/17

y^((2)) +y=sec^3 x

$${y}^{\left(\mathrm{2}\right)} +{y}=\mathrm{sec}\:^{\mathrm{3}} {x} \\ $$

Commented by prakash jain last updated on 31/Dec/17

g(x)=sec^3 x Homogenous solution λ^2 +1=0⇒λ=±i y_h =c_1 e^(ix) +c_2 e^(−ix) y_h =(c_1 +c_2 )cos x+i(c_1 −c_2 )cos x y_h =C_1 cos x+C_2 cos x y_1 (x)=cos x y_2 (x)=sin (x) W(y_1 ,y_2 )= determinant (((cos x),(sin x)),((−sin x),(cos x)))=1 u_1 (x)=−∫((y_2 (x)g(x))/1)dx =−∫sec^3 xsin xdx=−∫sec^2 xtan xdx =−(1/2)sec^2 x u_2 (x)=∫((y_1 (x)g(x))/1)dx =∫sec^3 x∙cos xdx=tan x y_p (x)=u_1 y_1 +u_2 y_2 =−((sec x)/2)+sin xtan x

$${g}\left({x}\right)=\mathrm{sec}^{\mathrm{3}} {x} \\ $$$$\mathrm{Homogenous}\:\mathrm{solution} \\ $$$$\lambda^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow\lambda=\pm{i} \\ $$$${y}_{{h}} ={c}_{\mathrm{1}} {e}^{{ix}} +{c}_{\mathrm{2}} {e}^{−{ix}} \\ $$$${y}_{{h}} =\left({c}_{\mathrm{1}} +{c}_{\mathrm{2}} \right)\mathrm{cos}\:{x}+{i}\left({c}_{\mathrm{1}} −{c}_{\mathrm{2}} \right)\mathrm{cos}\:{x} \\ $$$${y}_{{h}} ={C}_{\mathrm{1}} \mathrm{cos}\:{x}+{C}_{\mathrm{2}} \mathrm{cos}\:{x} \\ $$$${y}_{\mathrm{1}} \left({x}\right)=\mathrm{cos}\:{x} \\ $$$${y}_{\mathrm{2}} \left({x}\right)=\mathrm{sin}\:\left({x}\right) \\ $$$$\mathscr{W}\left({y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \right)=\begin{vmatrix}{\mathrm{cos}\:{x}}&{\mathrm{sin}\:{x}}\\{−\mathrm{sin}\:{x}}&{\mathrm{cos}\:{x}}\end{vmatrix}=\mathrm{1} \\ $$$${u}_{\mathrm{1}} \left({x}\right)=−\int\frac{{y}_{\mathrm{2}} \left({x}\right){g}\left({x}\right)}{\mathrm{1}}{dx} \\ $$$$=−\int\mathrm{sec}^{\mathrm{3}} {x}\mathrm{sin}\:{xdx}=−\int\mathrm{sec}^{\mathrm{2}} {x}\mathrm{tan}\:{xdx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} {x} \\ $$$${u}_{\mathrm{2}} \left({x}\right)=\int\frac{{y}_{\mathrm{1}} \left({x}\right){g}\left({x}\right)}{\mathrm{1}}{dx} \\ $$$$=\int\mathrm{sec}^{\mathrm{3}} {x}\centerdot\mathrm{cos}\:{xdx}=\mathrm{tan}\:{x} \\ $$$${y}_{{p}} \left({x}\right)={u}_{\mathrm{1}} {y}_{\mathrm{1}} +{u}_{\mathrm{2}} {y}_{\mathrm{2}} =−\frac{\mathrm{sec}\:{x}}{\mathrm{2}}+\mathrm{sin}\:{x}\mathrm{tan}\:{x} \\ $$

Answered by prakash jain last updated on 31/Dec/17

y=y_h +y_p

$${y}={y}_{{h}} +{y}_{{p}} \\ $$

Commented by sorour87 last updated on 31/Dec/17

ok. thanks

$${ok}.\:{thanks} \\ $$

Commented by prakash jain last updated on 31/Dec/17

The following page desxribes the method of variation of parameters. http://tutorial.math.lamar.edu/Classes/DE/VariationofParameters.aspx

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