Answer to Question #278268 in Differential Equations for Jericho

Question

0.4i'+10i=40-sin35t

Answer 1

$$0.4i'+10i=40-\sin35t$$ $$i'+25i=100-2.5\sin35t$$

Integrating factor

$$\mu(t)=e^{\int25dt}=e^{25t}$$

$$e^{25t}(i'+25i)=100e^{25t}-2.5e^{25t}\sin35t$$

$$d(e^{25t}i)=(100e^{25t}-2.5e^{25t}\sin35t)dt$$

Integrate

$$\int d(e^{25t}i) =\int (100e^{25t}-2.5e^{25t}\sin35t)dt$$

$$I_1=\int e^{25t}\sin35tdt$$

$$u=\sin35t, du=35\cos35tdt$$

$$dv=e^{25t}dt, v=\int e^{25t}dt=0.04e^{25t}$$

$$I_1=\int e^{25t}\sin35tdt$$

$$=0.04e^{25t}\sin35t-1.4\int e^{25t}\cos35tdt$$

$$\int e^{25t}\cos35tdt$$

$$u=\cos35t, du=-35\sin35tdt$$

$$dv=e^{25t}dt, v=\int e^{25t}dt=0.04e^{25t}$$

$$\int e^{25t}\cos35tdt=0.04e^{25t}\cos35t$$ $$+1.4\int e^{25t}\sin35tdt$$

$$\int e^{25t}\sin35tdt=0.04e^{25t}\sin35t-$$

$$-0.056e^{25t}\cos35t-1.96\int e^{25t}\sin35tdt$$

$$2.96\int e^{25t}\sin35tdt=0.04e^{25t}\sin35t$$

$$-0.056e^{25t}\cos35t$$

$$\int e^{25t}\sin35tdt=\dfrac{2}{148}e^{25t}\sin35t-\dfrac{2.8}{148}e^{25t}\cos35t$$

$$e^{25t}i=4e^{25t}-\dfrac{5}{148}e^{25t}\sin35t+\dfrac{7}{148}e^{25t}\cos35t+C$$

$$i=4-\dfrac{5}{148}\sin35t+\dfrac{7}{148}\cos35t+Ce^{-25t}$$