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Question Number 1898 by 123456 last updated on 22/Oct/15

$$\frac{df}{dt}=\alpha f+\beta t+\gamma$$$$f\left(t\right)=??$$

Answered by Yozzy last updated on 22/Oct/15

$$\frac{df}{dt}=\alpha f+\beta t+\gamma whereIassumethat\alpha ,\beta ,\gamma areconstants.Thisequationmaybe$$$$rewrittenas\frac{df}{dt}-\alpha f=\beta t+\gamma (\ast ).Theequationisafirstorderlinearnon-homogeneous$$$$differentialequationwhichmaybesolvedbyuseofanintegratingfactor\psi \left(t\right).$$$$Let\psi \left(t\right)=e^{\int -\alpha dt}=e^{-\alpha t}.Multiplyingbothsidesof(\ast )by\psi \left(t\right)weget$$$$\psi \left(t\right)\frac{df}{dt}+\psi \left(t\right)(-\alpha )f=\psi \left(t\right)(\beta t+\gamma )$$$$\Rightarrow e^{-\alpha t}\frac{df}{dt}+(-\alpha e^{-\alpha t})f=e^{-\alpha t}(\beta t+\gamma )$$$$Noticethat\frac{d}{dt}\left(e^{-\alpha t}f\right)=e^{-\alpha t}\frac{df}{dt}+(-\alpha e^{-\alpha t})f.$$$$\therefore \frac{d}{dt}\left({fe}^{-\alpha t}\right)=e^{-\alpha t}(\beta t+\gamma )(\ast \ast )$$$$Integratingbothsidesof(\ast \ast )weget$$$${fe}^{-\alpha t}=\int e^{-\alpha t}(\beta t+\gamma )dt$$$$f=e^{\alpha t}\int e^{-\alpha t}(\beta t+\gamma )dt$$$$Byintegratingbyparts,$$$$\int e^{-\alpha t}(\beta t+\gamma )dt=\frac{e^{-\alpha t}}{-\alpha }(\beta t+\gamma )-\int \frac{e^{-\alpha t}}{-\alpha }\times \beta dt(\alpha \ne 0)$$$$=\frac{e^{-\alpha t}(\beta t+\gamma )}{-\alpha }+\frac{\beta }{\alpha }\times \frac{e^{-\alpha t}}{-\alpha }+D$$$$\int e^{-\alpha t}(\beta t+\gamma )dt=\frac{e^{-\alpha t}}{-\alpha }(\beta t+\gamma +\frac{\beta }{\alpha })+D$$$$\therefore f=e^{\alpha t}(\frac{e^{-\alpha t}}{-\alpha }(\beta t+\frac{\beta }{\alpha }+\gamma )+D)$$$$f\left(t\right)={De}^{\alpha t}-\frac{1}{\alpha }\times \frac{\beta \alpha t+\beta +\alpha \gamma }{\alpha }$$$$f\left(t\right)={De}^{\alpha t}-\frac{\beta \alpha t+\beta +\alpha \gamma }{{\alpha }^2}(\alpha \ne 0)whereDisanarbitraryconstant.$$$$CheckingSolution:$$$$\frac{df}{dt}=\alpha {De}^{\alpha t}-\frac{\beta }{\alpha }$$$${De}^{\alpha t}=f+\frac{\beta }{\alpha }t+\frac{\beta +\alpha \gamma }{{\alpha }^2}$$$$\Rightarrow \frac{df}{dt}=\alpha f+\beta t+\alpha \frac{\beta }{{\alpha }^2}+\alpha \times \frac{\alpha \gamma }{{\alpha }^2}-\frac{\beta }{\alpha }$$$$\frac{df}{dt}=\alpha f+\beta t+\gamma asgiven.$$$$$$

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