Solve the given differential equation. If an initial condition is

compagnia04 2021-12-29 Answered
Solve the given differential equation. If an initial condition is given, also find the solution that satisfies it.
(ex+1)dydx=yyex
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Expert Answer

Edward Patten
Answered 2021-12-30 Author has 38 answers
Step 1
Given differential equation,
(ex+1)dydx=yyex
Step 2
Since we have,
(ex+1)dydx=yyex
(ex+1)dydx=y(1ex)
dyy=1ex1+exdx [Variable seperable form]
Integrate both sides,
dyy=1ex1+exdx (1)
Step 3
Now firstly simplify 1ex1+exdx,
Substitute ex=uexdx=dudx=1exdu, so
1ex1+exdx=1u1+uduu
=1uu(1u)du (2)
Simplify 1uu(1+u) using partial fraction method,
1uu(1+u)=Au+B1+u (3)
1u=A(1+u)+Bu
put u=0,A=1
Put u=1,B=2
Substitute the value of A and B in (3),
1uu(1+u)=1u21+u (4)
Using (4) i (2),
1ex1+ex=(1u21+u)du
=lnu2ln
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Mollie Nash
Answered 2021-12-31 Author has 33 answers
Given differential equation is (ex+1)dydx=yyex
(ex+1)dydx=y(1ex)
on separaing the variables, we get
dyy=1ex1+exdx
Integrating on both sides
dyy=1ex1+exdx
lny=(1+ex)2ex1+exdx
lny=(12ex1+ex)dx
lny=x2ln(1+ex)+c
y=ex2h(1+e)+c
y=exe2h(1+e)ec
y=c1exeln(1+e) where c1=ec
y=c1ex(1+ex)2
Therefore the solution of the given differential equation is y=c1ex(1+ex)2
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karton
Answered 2022-01-09 Author has 439 answers

(ex+1)dydx=yyex(ex+1)dydx=yyex(ex+1)dydx=y(1ex)dyy=1ex1+exdxdyy=1ex1+exdx(1)1ex1+exdx,ex=uexdx=dudx=1exdu,1ex1+exdx=1u1+uduu=1uu(1u)du(2)1uu(1+u),1uu(1+u)=Au+B1+u(3)1u=A(1+u)+Buu=0,A=1u=1,B=21uu(1+u)=1u21+u (4)Using (4) and (2),1ex1+ex=(1u21+u)du=lnu2ln(1+u)+c=ln(ex)2ln(1+ex)+c1ex1+exdx in (1),dyy=ln(ex)2ln(1+ex)+clny=x2ln(1+ex)+clny=(x+c)+ln(1+ex)2y=ex+c(1+ex)2Answer: y=ex+c(1+ex)2

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