Solve the differential equation

[fresh_42]

kindly check my post 18...i think it was just fine i have seen the error in typing $(1-v^2)$ in post 15... instead of $(1-v)$...corrected in post 18...

Close but wrong. We have $\displaystyle{\int} \dfrac{dx}{x}= \ln x$ which you got right. But the other one is
$$
\int \dfrac{1+v}{1-v} \,dv = \int \dfrac{dv}{1-v} + \int \dfrac{v}{1-v}\,dv = - \ln |1-v| +\, ?
$$
The second integral doesn't result in a summand $1-v$, which is what you got wrong. You can either solve $\displaystyle{\int \dfrac{v}{1-v}\,dv}$ by hand (integration by parts) or look it up in a table, e.g. Wikipedia, or check it with WolframAlpha.

 

[chwala]

Close but wrong. We have $\displaystyle{\int} \dfrac{dx}{x}= \ln x$ which you got right. But the other one is
$$
\int \dfrac{1+v}{1-v} \,dv = \int \dfrac{dv}{1-v} + \int \dfrac{v}{1-v}\,dv = - \ln |1-v| +\, ?
$$
The second integral doesn't result in a summand $1-v$, which is what you got wrong. You can either solve $\displaystyle{\int \dfrac{v}{1-v}\,dv}$ by hand (integration by parts) or look it up in a table, e.g. Wikipedia, or check it with WolframAlpha.

therefore, are you implying that if i use substitution,
$u=1-v, v= 1-u, dv=-du, y(1)=0$
$-∫\frac {2-u}{u} du$=∫$\frac {dx} {x}$
$-2ln u+u-ln x=k$
using the initial condition $y(1)=0, k=1$
therefore,
$-2ln (1-v)+(1-v)-ln x=1$
$1-\frac {y}{x}$=$1+ln x+ln (1-\frac {y}{x})^2$
ok...i will look at this again...i always learn math by making mistakes..

 

[Mark44]

therefore, are you implying that if i use substitution,
$u=1-v, v= 1-u, dv=-du, y(1)=0$

The initial condition is completely irrelevant here. It shouldn't play a role until after you have solved the differential equation in v and x, and then replaced v with y/x.

$-∫\frac {2-u}{u} du$=∫$\frac {dx} {x}$
$-2ln u+u-ln x=k$
using the initial condition $y(1)=0, k=1$
therefore,
$-2ln (1-v)+(1-v)-ln x=1$
$1-\frac {y}{x}$=$1+ln x+ln (1-\frac {y}{x})^2$
ok...i will look at this again...i always learn math by making mistakes..

Several posts back you gave the book's solution. Does your solution here match that one? You can always verify that a solution you find is actually a solution by 1) confirming that your solution matches the initial condition, and 2) by confirming that you can differentiate your solution to get back to the given differential equation.

 

[chwala]

The initial condition is completely irrelevant here. It shouldn't play a role until after you have solved the differential equation in v and x, and then replaced v with y/x.
Several posts back you gave the book's solution. Does your solution here match that one? You can always verify that a solution you find is actually a solution by 1) confirming that your solution matches the initial condition, and 2) by confirming that you can differentiate your solution to get back to the given differential equation.

i think the textbook answer might be wrong unless i am missing something, let me post all my steps now. Further, i think the initial condition is necessary so as to arrive at a solution. give me a moment i post all my steps...

 

[chwala]

now we have,
∫$\frac {1+v}{1-v} dv = ∫\frac {dx}{x}$
Let $u=1-v$, →$du= -dv$ →$v=1-u$
-∫$\frac {2-u}{u} du= ∫\frac {dx}{x}$
∫$(1-2u^-{1}) du= ∫\frac {dx}{x}$
$u-2ln u-ln x=k$
on using $y(1)=0, k=1$
$(1-v)-ln(u)^2-ln x=1$
1-$\frac {y}{x}-ln(1-yx^-1)^2-ln x=1$
-$\frac {y}{x}= ln (1-yx^-1)^2+ln x$
-$\frac {y}{x}=ln x [\frac {x^2-2xy+y^2}{x^2}$]
$-y=xln [\frac {x^2-2xy+y^2}{x}$]
$y=-xln [\frac {x^2-2xy+y^2}{x}$]
now the above is my steps...

the textbook solution is
$y=x ln [\frac {x}{x^2-2xy+y^2}]$

 

[fresh_42]

i think the textbook answer might be wrong unless i am missing something, let me post all my steps now. Further, i think the initial condition is necessary so as to arrive at a solution. give me a moment i post all my steps...

The textbook answer is correct. What do you get for $\displaystyle{\int \dfrac{v}{1-v}\,dv}\;?$
@Mark44 has told you the next steps:

Solve the integrals correctly! With constant term(s).
Undo the $y \to v$ substitution.
Apply the boundary conditions to specify the constants from integration.
Algebraically modify the result in order to get the form of the textbook answer.

But first solve the integral above.

 

[chwala]

The textbook answer is correct. What do you get for $\displaystyle{\int \dfrac{v}{1-v}\,dv}\;?$
@Mark44 has told you the next steps:

Solve the integrals correctly! With constant term(s).
Undo the $y \to v$ substitution.
Apply the boundary conditions to specify the constants from integration.
Algebraically modify the result in order to get the form of the textbook answer.

But first solve the integral above.

Bingo!
I just checked the two solutions! I got it, they're the same... Am in a meeting, by using the basic laws of logs, my solution can be expressed as the other. Therefore, I was just right. A quick way of proving is just by plugging arbitrary values for x and y in the two solutions!

 

[WWGD]

The textbook answer is correct. What do you get for $\displaystyle{\int \dfrac{v}{1-v}\,dv}\;?$
@Mark44 has told you the next steps:

Solve the integrals correctly! With constant term(s).
Undo the $y \to v$ substitution.
Apply the boundary conditions to specify the constants from integration.
Algebraically modify the result in order to get the form of the textbook answer.

But first solve the integral above.

I think his is correct: $-ln(b/a)=-( lnb-lna)=lna-lnb=ln(a/b).$

 

[fresh_42]

I think his is correct: $-ln(b/a)=-( lnb-lna)=lna-lnb=ln(a/b).$

That was not the problem. He wrote, at least the first time, that $\displaystyle{\int \dfrac{v}{1-v}}\,dv = -(1-v)-\ln (1-v)$ which is wrong on several levels: $\displaystyle{\int \dfrac{v}{1-v}}\,dv = -v-\ln |1-v| + C$

But he edits his posts all the time and he doesn't always tell what he actually did, that I lost track. His first answer to the integral was wrong and if at all, he corrected it via edit, which makes the thread almost unreadable.

 

[chwala]

That was not the problem. He wrote, at least the first time, that $\displaystyle{\int \dfrac{v}{1-v}}\,dv = -(1-v)-\ln (1-v)$ which is wrong on several levels: $\displaystyle{\int \dfrac{v}{1-v}}\,dv = -v-\ln |1-v| + C$

But he edits his posts all the time and he doesn't always tell what he actually did, that I lost track. His first answer to the integral was wrong and if at all, he corrected it via edit, which makes the thread almost unreadable.

Sorry, thanks for your insight though. I thought my post number 35 was clear. I even indicated that I would show all my steps,...
I've noted your concerns on editing posts. I will desist from it to avoid future confusion...

 

[WWGD]

Sorry, thanks for your insight though. I thought my post number 35 was clear. I even indicated that I would show all my steps,...
I've noted your concerns on editing posts. I will desist from it to avoid future confusion...

Or write in the post the fact that you are editing as well as the details of the edit.

 

[fresh_42]

Or write in the post the fact that you are editing as well as the details of the edit.

A new post is better, especially if there are already answers! And changing calculations forces readers to read the same calculation again and again. That will not work on the long run.

 

[fresh_42]

I was referring to this post:

ok let me solve it, am now getting after simplification,
$\frac {1+v} {1−v}$ dv= $\frac {dx} {x}$
on integration i am getting;
$2ln(1−v)−(1−v)=lnx$
→ $2ln(1−v)− ln x= (1-v)$

The integration is wrong here. It must not be $-(1-v)$ in the linear term, it has to be $-v$.

 

[haruspex]

@chwala , it might help you in future with such problems to see how the substitution y=vx can be spotted. One thing I check up front with a ODE is whether there is a consistent way of assigning dimension to the variables. In the present case, giving x and y the same dimension works out, suggesting solutions like y=Cx, for some constant. Indeed, C=1 is a solution.
To find more, you can look for variants such as y=xv (v having no dimension).
Had the original equation been dimensionally satisfied by y=x², you might have tried y=x²v, etc.

 

[chwala]

Thanks haruspex, fresh and wwgd on your input on this problem. I have learned something on the approach... i guess i have to take time and avoid silly mistakes and math errors...