Show that the differential equation has no solution that satisfies g(0) = 1

mathmari

Well-known member
MHB Site Helper
hello! I am facing some difficulties at the following exercise. "show that $$g(x)=x \cdot F(x)$$, where $$F(x)=\int_{0}^{x} {s(x)}dt$$, $$s(x)=\frac{sin(x)}{x}$$, satisfies the diffential equation $$xy'(x)-y(x)=xsin(x)$$, x ε R, and find all the solutions in this space. Show that the differential equation has no solution that satisfies g(0)=1. "
I have shown that g(x) satisfies the equation by replacing y with g. Then I found that all the solutions are $$y(x)=x(c+F(x))$$. Is this right so far???
How can I show that the differential equation has no solution that satisfies g(0)=1??

chisigma

Well-known member
Re: Solution of differential equation

hello! I am facing some difficulties at the following exercise. "show that $$g(x)=x \cdot F(x)$$, where $$F(x)=\int_{0}^{x} {s(x)}dt$$, $$s(x)=\frac{sin(x)}{x}$$, satisfies the diffential equation $$xy'(x)-y(x)=xsin(x)$$, x ε R, and find all the solutions in this space. Show that the differential equation has no solution that satisfies g(0)=1. "
I have shown that g(x) satisfies the equation by replacing y with g. Then I found that all the solutions are $$y(x)=x(c+F(x))$$. Is this right so far???
How can I show that the differential equation has no solution that satisfies g(0)=1??
Let's write the ODE in the form...

$\displaystyle y^{\ '} = \frac{y}{x} + \sin x\ (1)$

... and the 'standard approach' leads to the solution...

$\displaystyle y = e^{\int \frac{d x}{x}}\ \{\int \sin x\ e^{- \int \frac{d x}{x}}\ dx + c \} = x\ \{ \int \frac{sin x}{x}\ dx + c \} = c\ x + x\ \text{Si}\ (x)\ (2)$

Because in any case is y(0)= 0, no solution exist for the initial condition y(0) =1...

Kind regards

$\chi$ $\sigma$

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

Thank you! I have also an other question
How could we explain that the fact that no solution exist for the initial condition y(0)=1 doesn't affect the existence of solution theorem ?

chisigma

Well-known member
Re: Solution of differential equation

Thank you! I have also an other question
How could we explain that the fact that no solution exist for the initial condition y(0)=1 doesn't affect the existence of solution theorem ?
An ODE in the form...

$\displaystyle y^{\ '} = f(x,y),\ y(x_{0})= y_{0}\ (1)$

... admits solution in a neighbourhood of $(x_{0},y_{0})$ only if f(x,y) and its first order partial derivatives are continuos in $(x_{0},y_{0})$. In Your case is $\displaystyle f(x,y)= \frac{y}{x} + \sin x$ and this condition isn't verified in $(0,1)$...

Kind regards

$\chi$ $\sigma$

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

So you mean that the theorem is not verified, and that there is no solution? Or do I understand it wrong?

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chisigma

Well-known member
Re: Solution of differential equation

So you mean that the theorem is not verified, and that there is no solution? Or do I understand it wrong?
You understand exactly!... a solution exists for all initial conditions $y(x_{0})=y_{0}$ provided that $x_{0} \ne 0$...

Kind regards

$\chi$ $\sigma$

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

For the exercise I have to show that the fact that no solution exist for the initial condition y(0)=1 does not contradict the theorem. Could you give me a hint how to do this?

Klaas van Aarsen

MHB Seeker
Staff member
Re: Solution of differential equation

Hey mathmari!

For the exercise I have to show that the fact that no solution exist for the initial condition y(0)=1 does not contradict the theorem. Could you give me a hint how to do this?
Can you quote the existence theorem?
What are its conditions?
Are they all satisfied?
If not all conditions are satisfied, this example cannot contradict the theorem since the theorem won't be applicable.

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

The existence theorem is:
Let$$y'+p(x)y=g(x)$$, $$y(x_{0})=y_{0}$$ be a first order linear differential equation such that $$p(x)$$ and $$g(x)$$ are both continuous on an open interval $$a<x<b$$ and the interval contains $$x_{0}$$. Then there is a unique solution on that interval.

At the exercise is $$x_{0}=0$$, but $$p(x)=-\frac{1}{x}$$ isn't continuous on an open interval that contains $$0$$. Since this condition isn't satisfied, what does this mean?

Klaas van Aarsen

MHB Seeker
Staff member
Re: Solution of differential equation

The existence theorem is:
Let$$y'+p(x)y=g(x)$$, $$y(x_{0})=y_{0}$$ be a first order linear differential equation such that $$p(x)$$ and $$g(x)$$ are both continuous on an open interval $$a<x<b$$ and the interval contains $$x_{0}$$. Then there is a unique solution on that interval.

At the exercise is $$x_{0}=0$$, but $$p(x)=-\frac{1}{x}$$ isn't continuous on an open interval that contains $$0$$. Since this condition isn't satisfied, what does this mean?
Good!

It means that the theorem is not applicable since its preconditions are not satisfied.

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

The exercise asks me to explain why the fact that there is no solution that satisfies f(0)=1 doesn't contradict the existence theorem. So is the answer that theorem isn't applicable?

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

Or is the solution $$f(0)=1$$ maybe a peculiar solution, so the existence theorem can be applicated for $$x \neq 0$$???

chisigma

Well-known member
Re: Solution of differential equation

Or is the solution $$f(0)=1$$ maybe a peculiar solution, so the existence theorem can be applicated for $$x \neq 0$$???
In...

http://mathhelpboards.com/different...ion-differential-equation-7571.html#post34482

... it has been demonstrated that the general solution of the ODE...

$\displaystyle y^{\ '} = \frac{y}{x} + \sin x\ (1)$

... is...

$\displaystyle y(x) = c\ x + x\ \text{Si}\ (x)\ (2)$

Now observing (2) You realize that, no matter which is c, is y(0)=0, so that You cannot impose in x=0 other values that y=0... but if You impose $y(0)=0$ what of the infinite values of c gives us the 'right solution'?...

Kind regards

$\chi$ $\sigma$

Klaas van Aarsen

MHB Seeker
Staff member
Re: Solution of differential equation

Or is the solution $$f(0)=1$$ maybe a peculiar solution, so the existence theorem can be applicated for $$x \neq 0$$???
Here's my take on the problem.

No, $$f(0)=1$$ is not a peculiar solution.
The existance theorem, as you state it, is not applicable, so that does not tell you if there is a solution for $$f(0)=1$$.
However, filling in the boundary criterium into the differential equation tells you immediately that there cannot be a solution, since the equation is not satisfied.

mathmari

Well-known member
MHB Site Helper
Re: Solution of differential equation

Thank you!!!