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Question Number 157538 by ghakhan88 last updated on 24/Oct/21

Commented by ghakhan88 last updated on 24/Oct/21

$$Iwishsomeonecouldhelpme$$

Answered by metamorfose last updated on 25/Oct/21

$$(\bullet )u\left(x\right)=x+1+{\int }_0^x\frac{t^{a-1}}{x^a}u\left(t\right)dt$$$$derivate(\bullet )$$$$u^{\prime }\left(x\right)=1+\frac{u\left(x\right)}{x}-\frac{a}{x}{\int }_0^x\frac{t^{a-1}}{x^a}u\left(t\right)dt(\bullet \bullet )$$$$(\bullet )give:{\int }_0^x\frac{t^{a-1}}{x^a}u\left(t\right)dt=u\left(x\right)-(x+1)$$$$hence:$$$$u^{\prime }\left(x\right)+\frac{a-1}{x}u\left(x\right)=1-a-\frac{a}{x}\left(ED\right)$$$$\left(ED\right)isadifferentialequation$$$$type{y}^{\prime }\left(t\right)+a\left(t\right)y\left(t\right)=b\left(t\right)...$$$$$$

Commented by ghakhan88 last updated on 27/Oct/21

$$Itriedtosolvetheeq\left(ED\right)usingLaplace,andIgetronganswer!$$

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