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Question Number 27028 by sorour87 last updated on 01/Jan/18

y^((2)) +4y=sinh x×sin 2x

$${y}^{\left(\mathrm{2}\right)} +\mathrm{4}{y}=\mathrm{sinh}\:{x}×\mathrm{sin}\:\mathrm{2}{x} \\ $$

Answered by prakash jain last updated on 02/Jan/18

λ^2 +4=0  λ=±2i  y_h =c_1 cos 2x+c_2 sin 2x  y_1 =cos 2x,y_2 =sin 2x  W= determinant (((cos 2x),(sin 2x)),((−2sin 2x),(2cos 2x)))=2  v_1 =−∫((sinh x×sin 2x)/2)×sin 2xdx  v_1 =−(1/2)∫(1/2)(1−cos 4x)×sinh xdx  =−(1/4)∫sinh x−∫cos 4xsinh xdx  v_2 =∫((sinh x×sin 2x)/2)×cos  2xdx  y_p =v_1 y_1 +v_2 y_2   y=y_h +y_p

$$\lambda^{\mathrm{2}} +\mathrm{4}=\mathrm{0} \\ $$$$\lambda=\pm\mathrm{2}{i} \\ $$$${y}_{{h}} ={c}_{\mathrm{1}} \mathrm{cos}\:\mathrm{2}{x}+{c}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}{x} \\ $$$${y}_{\mathrm{1}} =\mathrm{cos}\:\mathrm{2}{x},{y}_{\mathrm{2}} =\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\mathscr{W}=\begin{vmatrix}{\mathrm{cos}\:\mathrm{2}{x}}&{\mathrm{sin}\:\mathrm{2}{x}}\\{−\mathrm{2sin}\:\mathrm{2}{x}}&{\mathrm{2cos}\:\mathrm{2}{x}}\end{vmatrix}=\mathrm{2} \\ $$$${v}_{\mathrm{1}} =−\int\frac{\mathrm{sinh}\:{x}×\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}×\mathrm{sin}\:\mathrm{2}{xdx} \\ $$$${v}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{cos}\:\mathrm{4}{x}\right)×\mathrm{sinh}\:{xdx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sinh}\:{x}−\int\mathrm{cos}\:\mathrm{4}{x}\mathrm{sinh}\:{xdx} \\ $$$${v}_{\mathrm{2}} =\int\frac{\mathrm{sinh}\:{x}×\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}×\mathrm{cos}\:\:\mathrm{2}{xdx} \\ $$$${y}_{{p}} ={v}_{\mathrm{1}} {y}_{\mathrm{1}} +{v}_{\mathrm{2}} {y}_{\mathrm{2}} \\ $$$${y}={y}_{{h}} +{y}_{{p}} \\ $$

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