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Question: If \(u = \int_{}^{}e^{ax}\cos bxdx\) and\(v = \int_{}^{}e^{ax}\sin bxdx,\) then \[(a^{2} + b^{2})(u...

If u=∫eaxcos⁡bxdxu = \int_{}^{}e^{ax}\cos bxdx andv=∫eaxsin⁡bxdx,v = \int_{}^{}e^{ax}\sin bxdx, then

(a2+b2)(u2+v2)=(a^{2} + b^{2})(u^{2} + v^{2}) =

A

2eax2e^{ax}

B

(a2+b2)e2ax(a^{2} + b^{2})e^{2ax}

C

e2axe^{2ax}

D

(a2−b2)e2ax(a^{2} - b^{2})e^{2ax}

Answer

e2axe^{2ax}

Explanation

Solution

u=∫eaxcos⁡bxdx\mathbf{u =}\int_{}^{}{\mathbf{e}^{\mathbf{ax}}\mathbf{\cos}\mathbf{b}\mathbf{xdx}} =eaxsin⁡bxb−ab∫eax.bxdx=eaxsin⁡bxb−abv⇒bu+av=eaxsin⁡bx\mathbf{=}\mathbf{e}^{\mathbf{ax}}\frac{\mathbf{\sin}\mathbf{b}\mathbf{x}}{\mathbf{b}}\mathbf{-}\frac{\mathbf{a}}{\mathbf{b}}\int_{}^{}{\mathbf{e}^{\mathbf{ax}}\mathbf{.bxdx =}\frac{\mathbf{e}^{\mathbf{ax}}\mathbf{\sin}\mathbf{b}\mathbf{x}}{\mathbf{b}}}\mathbf{-}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{v} \Rightarrow bu + av = e^{ax}\sin bx ......(i)

Similarly, bv−au=−eaxcos⁡bxbv - au = - e^{ax}\cos bx .....(ii)

Squaring (i) and (ii) and adding. We get,

(a2+b2)(u2+v2)=e2ax(a^{2} + b^{2})(u^{2} + v^{2}) = e^{2ax}.