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Question Number 74218 by malikmasood3535@gmail.com last updated on 20/Nov/19

$$verifythaty\left(x\right)=e^x(\cos e^x-e^x\sin e^x)isthesolutionofintegralequationy\left(x\right)=(1-{xe}^{2x})\cos 1-e^{2x}\sin 1+\underset{0}{\stackrel{x}{\int }}\{1-(x-t)e^{2x}\}y\left(t\right)dt$$

Answered by mind is power last updated on 20/Nov/19

$$y\left(x\right)=(1-{xe}^{2x})cos\left(1\right)-e^{2x}sin\left(1\right)+{\int }_0^x(1-(x-t)e^{2x}\}y\left(t\right)dt$$$$y\left(0\right)=cos\left(1\right)-sin\left(1\right)$$$$g\left(x\right)={\int }_0^x(1-(x-t)e^{2x}y\left(t\right)dt$$$$g\left(x\right)=(1-{xe}^{2x}){\int }_0^{x}y\left(t\right)dt+e^{2x}{\int }_0^{x}ty\left(t\right)dt$$$$fory\left(t\right)=e^t(cos\left(e^t\right)-e^{t}son\left(e^t\right))$$$${\int }_0^x{e}^t\left(cos\left(e^t\right)\right)-e^{2t}sin\left(e^t\right)dt=e^{x}cos\left(e^x\right)-cos\left(1\right)$$$${\int }_0^{x}t.(e^t(cos\left(e^t\right)-e^{t}sin\left(e^t\right))dtbypart$$$$={[t.e^{t}cos\left(e^t\right)]}_0^x-\int e^{t}cos\left(e^t\right)dt$$$$={xe}^{x}cos\left(e^x\right)-\left[{sine}^t\right]={xe}^{x}cos\left(e^x\right)-{sine}^x+sin\left(1\right)$$$$g\left(x\right)=(1-{xe}^{2x})(e^{x}cos\left(e^x\right)-cos\left(1\right))+e^{2x}({xe}^{x}cos\left(e^x\right)-sin\left(e^x\right)+sin\left(1\right))$$$$=-cos\left(1\right)+{xe}^{2x}cos\left(1\right)+e^{x}cos\left(e^x\right)-e^{2x}sin\left(e^x\right)+e^{2x}sin\left(1\right)$$$$(1-{xe}^{2x})cos\left(1\right)-e^{2x}sin\left(1\right)+g\left(x\right)$$$$=e^{x}cos\left(e^x\right)-e^{2x}sin\left(e^x\right)=y\left(x\right)$$$$\Rightarrow y\left(x\right)e^x(cos\left(e^x\right)-e^{x}sin\left(e^x\right))issolution$$$$$$$$$$

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