Gregory Jones

2021-12-31

Solve the differential equations $\left(t-1\right)3d\frac{s}{dt}+4\left(t-1\right)2s=t+1,t>1$

censoratojk

Step 1
Given:
Step 2
Calculation:
${\left(t-1\right)}^{3}\frac{ds}{dt}+4{\left(t-1\right)}^{2}s=t+1;t>1$
$\frac{ds}{dt}+\frac{4{\left(t-1\right)}^{2}}{{\left(t-1\right)}^{3}}s=\frac{t+1}{{\left(t-1\right)}^{3}}$

Compare the differential equation with $\frac{dy}{dx}+Py=Q$
Here
Solution of given differential equation is $y.\left(I.F\right)=\int Q.\left(I.F\right)dx+c$
Now, $s.{\left(t-1\right)}^{4}=\int \frac{t+1}{{\left(t-1\right)}^{3}}.{\left(t-1\right)}^{4}dt+c$
$s.{\left(t-1\right)}^{4}=\int \left(t+1\right).\left(t-1\right)dt+c$
$s.{\left(t-1\right)}^{4}=\int \left({t}^{2}-1\right)dt+c$
$s.{\left(t-1\right)}^{4}=\frac{{t}^{3}}{3}-t+c$

godsrvnt0706

${\left(t-1\right)}^{3}\frac{ds}{dt}+4{\left(t-1\right)}^{2}s=t+1;t>1$
$\frac{ds}{dt}+\frac{4{\left(t-1\right)}^{2}}{{\left(t-1\right)}^{3}}s=\frac{t+1}{{\left(t-1\right)}^{3}}$

$\frac{dy}{dx}+Py=Q$

$y.\left(I.F\right)=\int Q.\left(I.F\right)dx+c$
$s.{\left(t-1\right)}^{4}=\int \frac{t+1}{{\left(t-1\right)}^{3}}.{\left(t-1\right)}^{4}dt+c$
$s.{\left(t-1\right)}^{4}=\int \left(t+1\right).\left(t-1\right)dt+c$
$s.{\left(t-1\right)}^{4}=\int \left({t}^{2}-1\right)dt+c$
$s.{\left(t-1\right)}^{4}=\frac{{t}^{3}}{3}-t+c$

karton

Solution:
$\left(t-1{\right)}^{3}\frac{ds}{dt}+4\left(t-1{\right)}^{2}s=t+1;t>1$
$\frac{ds}{dt}+\frac{4\left(t-1{\right)}^{2}}{\left(t-1{\right)}^{3}}s=\frac{t+1}{\left(t-1{\right)}^{3}}$

Compare the differential equation with $\frac{dy}{dx}+Py=Q$

Solution of given differential equation is $y.\left(I.F\right)=\int Q.\left(I.F\right)dx+c$
$s.\left(t-1{\right)}^{4}=\int \frac{t+1}{\left(t-1{\right)}^{3}}.\left(t-1{\right)}^{4}dt+c$
$s.\left(t-1{\right)}^{4}=\frac{{t}^{3}}{3}-t+c$

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