Meldeaktezl

2022-09-03

Find a particular solution to

$${y}^{\u2033}-4{y}^{\prime}+4y=\frac{14{e}^{3t}}{{t}^{2}+1}$$

$${y}^{\u2033}-4{y}^{\prime}+4y=\frac{14{e}^{3t}}{{t}^{2}+1}$$

Ufumanaxi

Beginner2022-09-04Added 5 answers

The given differential equation is

$${y}^{\u2033}-4{y}^{\prime}+4y=\frac{14{e}^{2t}}{{t}^{2}+1}$$

The particular solution is

$${y}_{p}=\frac{1}{{D}^{2}-4D+4}(\frac{14{e}^{2t}}{{t}^{2}+1})$$ where $$D=\frac{d}{dx}$$

$$=\frac{1}{(D-2{)}^{2}}(\frac{14{e}^{2t}}{{t}^{2}+1})\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{(D+2-2{)}^{2}}\frac{1}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{{D}^{2}}\frac{1}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{D}{\mathrm{tan}}^{-1}(t)\phantom{\rule{0ex}{0ex}}=14{e}^{2t}[t{\mathrm{tan}}^{-1}(t)-\int \frac{t}{1+{t}^{2}}dt]\phantom{\rule{0ex}{0ex}}{y}_{p}=14{e}^{2t}[t{\mathrm{tan}}^{-1}-\frac{1}{2}\mathrm{log}(1+{t}^{2})]$$

$${y}^{\u2033}-4{y}^{\prime}+4y=\frac{14{e}^{2t}}{{t}^{2}+1}$$

The particular solution is

$${y}_{p}=\frac{1}{{D}^{2}-4D+4}(\frac{14{e}^{2t}}{{t}^{2}+1})$$ where $$D=\frac{d}{dx}$$

$$=\frac{1}{(D-2{)}^{2}}(\frac{14{e}^{2t}}{{t}^{2}+1})\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{(D+2-2{)}^{2}}\frac{1}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{{D}^{2}}\frac{1}{{t}^{2}+1}\phantom{\rule{0ex}{0ex}}=14{e}^{2t}\frac{1}{D}{\mathrm{tan}}^{-1}(t)\phantom{\rule{0ex}{0ex}}=14{e}^{2t}[t{\mathrm{tan}}^{-1}(t)-\int \frac{t}{1+{t}^{2}}dt]\phantom{\rule{0ex}{0ex}}{y}_{p}=14{e}^{2t}[t{\mathrm{tan}}^{-1}-\frac{1}{2}\mathrm{log}(1+{t}^{2})]$$

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