Solve the differential equations (t−1)3dsdt+4(t−1)2s=t+1,t&gt;1

Step 1 Given: (t−1)3dsdt+4(t−1)2 s=t+1;t&gt;1 Step 2 Calculation: (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 Compare the differential equation with dydx+Py=Q Here P=4(t−1)2 andQ=t+1(t−1)3 Solution of given differential equation is y.(I.F)=∫Q.(I.F)dx+c Now, s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=∫(t+1).(t−1)dt+c s.(t−1)4=∫(t2−1)dt+c s.(t−1)4=t33−t+c

(t−1)3dsdt+4(t−1)2 s=t+1;t&gt;1 (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 dydx+Py=Q P=4(t−1)2 andQ=t+1(t−1)3 y.(I.F)=∫Q.(I.F)dx+c s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=∫(t+1).(t−1)dt+c s.(t−1)4=∫(t2−1)dt+c s.(t−1)4=t33−t+c

Solution: (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 Compare the differential equation with dydx+Py=Q P=4(t−1)2 andQ=t+1(t−1)3 Solution of given differential equation is y.(I.F)=∫Q.(I.F)dx+c s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=t33−t+c

Step 1 Given: (t−1)3dsdt+4(t−1)2 s=t+1;t&gt;1 Step 2 Calculation: (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 Compare the differential equation with dydx+Py=Q Here P=4(t−1)2 andQ=t+1(t−1)3 Solution of given differential equation is y.(I.F)=∫Q.(I.F)dx+c Now, s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=∫(t+1).(t−1)dt+c s.(t−1)4=∫(t2−1)dt+c s.(t−1)4=t33−t+c

(t−1)3dsdt+4(t−1)2 s=t+1;t&gt;1 (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 dydx+Py=Q P=4(t−1)2 andQ=t+1(t−1)3 y.(I.F)=∫Q.(I.F)dx+c s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=∫(t+1).(t−1)dt+c s.(t−1)4=∫(t2−1)dt+c s.(t−1)4=t33−t+c

Solution: (t−1)3dsdt+4(t−1)2s=t+1;t&gt;1 dsdt+4(t−1)2(t−1)3s=t+1(t−1)3 dsdt+4(t−1) s=t+1(t−1)3 Compare the differential equation with dydx+Py=Q P=4(t−1)2 andQ=t+1(t−1)3 Solution of given differential equation is y.(I.F)=∫Q.(I.F)dx+c s.(t−1)4=∫t+1(t−1)3.(t−1)4dt+c s.(t−1)4=t33−t+c

Resolved: "Solve the differential equations (t−1)3dsdt+4(t−1)2s=t+1,t>1"

Gregory Jones

Answered question

2021-12-31

Solve the differential equations

(t - 1) 3 d \frac{s}{d t} + 4 (t - 1) 2 s = t + 1, t > 1

Answer & Explanation

censoratojk

Beginner2022-01-01Added 46 answers

Step 1
Given:

{(t - 1)}^{3} \frac{d s}{d t} + 4 {(t - 1)}^{2} s = t + 1; t > 1

Step 2
Calculation:

{(t - 1)}^{3} \frac{d s}{d t} + 4 {(t - 1)}^{2} s = t + 1; t > 1

\frac{d s}{d t} + \frac{4 {(t - 1)}^{2}}{{(t - 1)}^{3}} s = \frac{t + 1}{{(t - 1)}^{3}}

\frac{d s}{d t} + \frac{4}{(t - 1)} s = \frac{t + 1}{{(t - 1)}^{3}}

Compare the differential equation with

\frac{d y}{d x} + P y = Q

Here

P = \frac{4}{{(t - 1)}^{2}} and Q = \frac{t + 1}{{(t - 1)}^{3}}

Solution of given differential equation is

y . (I . F) = \int Q . (I . F) d x + c

Now,

s . {(t - 1)}^{4} = \int \frac{t + 1}{{(t - 1)}^{3}} . {(t - 1)}^{4} d t + c

s . {(t - 1)}^{4} = \int (t + 1) . (t - 1) d t + c

s . {(t - 1)}^{4} = \int (t^{2} - 1) d t + c

s . {(t - 1)}^{4} = \frac{t^{3}}{3} - t + c

godsrvnt0706

Beginner2022-01-02Added 31 answers

{(t - 1)}^{3} \frac{d s}{d t} + 4 {(t - 1)}^{2} s = t + 1; t > 1

{(t - 1)}^{3} \frac{d s}{d t} + 4 {(t - 1)}^{2} s = t + 1; t > 1

\frac{d s}{d t} + \frac{4 {(t - 1)}^{2}}{{(t - 1)}^{3}} s = \frac{t + 1}{{(t - 1)}^{3}}

\frac{d s}{d t} + \frac{4}{(t - 1)} s = \frac{t + 1}{{(t - 1)}^{3}}

\frac{d y}{d x} + P y = Q

P = \frac{4}{{(t - 1)}^{2}} and Q = \frac{t + 1}{{(t - 1)}^{3}}

y . (I . F) = \int Q . (I . F) d x + c

s . {(t - 1)}^{4} = \int \frac{t + 1}{{(t - 1)}^{3}} . {(t - 1)}^{4} d t + c

s . {(t - 1)}^{4} = \int (t + 1) . (t - 1) d t + c

s . {(t - 1)}^{4} = \int (t^{2} - 1) d t + c

s . {(t - 1)}^{4} = \frac{t^{3}}{3} - t + c

karton

Expert2022-01-10Added 613 answers

Solution:
$(t - 1)^{3} \frac{d s}{d t} + 4 (t - 1)^{2} s = t + 1; t > 1$
$\frac{d s}{d t} + \frac{4 (t - 1)^{2}}{(t - 1)^{3}} s = \frac{t + 1}{(t - 1)^{3}}$
$\frac{d s}{d t} + \frac{4}{(t - 1)} s = \frac{t + 1}{(t - 1)^{3}}$
Compare the differential equation with $\frac{d y}{d x} + P y = Q$
$P = \frac{4}{(t - 1)^{2}} and Q = \frac{t + 1}{(t - 1)^{3}}$
Solution of given differential equation is $y . (I . F) = \int Q . (I . F) d x + c$
$s . (t - 1)^{4} = \int \frac{t + 1}{(t - 1)^{3}} . (t - 1)^{4} d t + c$
$s . (t - 1)^{4} = \frac{t^{3}}{3} - t + c$

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