# Thread: First order non-linear ODE

1. ## First order non-linear ODE

Hi everyone,

x2y'-20+x2y2=0; y(1)=1

The question asks me to use substitution

y=1/u*du/dx to solve

So far I have

y'=-1/u2*du/dx+1/u*d2u/dx2

problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.

2. ## Re: First order non-linear ODE

Originally Posted by lillybeans
Hi everyone,

x2y'-20+x2y2=0; y(1)=1

The question asks me to use substitution

y=1/u*du/dx to solve

So far I have

y'=-1/u2*du/dx+1/u*d2u/dx2

problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.
That term should reduce out....

$\displaystyle y=\frac{u'}{u} \iff uy=u'$

$\displaystyle y'u+u'y=u'' \iff y'u+u'\frac{u'}{u}=u'' \iff y'=-\frac{(u')^2}{u^2}+u''$

If you sub this into the ode you get

$\displaystyle x^2\left( -\frac{(u')^2}{u^2}+u''\right)+x^2\left(\frac{u'}{u } \right)^2=20$

This gives

$\displaystyle u''(x)=\frac{20}{x^2}$

Can you finish from here?

3. ## Re: First order non-linear ODE

Originally Posted by lillybeans
Hi everyone,

x2y'-20+x2y2=0; y(1)=1

The question asks me to use substitution

y=1/u*du/dx to solve

So far I have

y'=-1/u2*du/dx+1/u*d2u/dx2

problem is when I sub in for the x2y2 term I'm going to get a derivative squared --> y2=(1/u*du/dx)2. Don't know how to resolve this. Please help.
After simplification, the terme (1/u*du/dx)² disapears

4. ## Re: First order non-linear ODE

Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x

For the first one I wrote:
yp=Px3+Qx2+Rx

Second one:
yp=(Px2+Qx+R)e4x

Neither of the answers are correct, according to the computer. Where did I go wrong?

5. ## Re: First order non-linear ODE

Originally Posted by lillybeans
Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x

For the first one I wrote:
yp=Px3+Qx2+Rx

Second one:
yp=(Px2+Qx+R)e4x

Neither of the answers are correct, according to the computer. Where did I go wrong?
I am not sure what method you are using to find the particular solutions, but here is one...

We can write the ODE as

$\displaystyle (D^3-9D^2+14D)y=x$

Factoring the differential operator we get

$\displaystyle D(D-7)(D-2)y=x$

So to make the equation equal to zero we act on both sides with $\displaystyle D^2$

$\displaystyle D^3(D-7)(D-2)y=0$

So $\displaystyle D^3$ annihilates

$\displaystyle y=c_1+c_2x+c_3x^2$

but we must get rid of the part from the complimentary solution (The constant term) so we get

$\displaystyle y=c_2x+c_3x^2$

Let me know if this works for you

6. ## Re: First order non-linear ODE

Thank you very much for you help, but unfortunately that didn't work. I even checked my answer using Wolfram and it seems to be correct, maybe something is wrong with web assignment.

I used the method of undetermined coefficients to find the particular solution.

7. ## Re: First order non-linear ODE

Originally Posted by lillybeans
Thank you both I solved this problem BUT I have another one which I am really stuck on this time, supposedly a much easier problem.

There are two questions that I am trying to solve on web assignment. The goal is to find a general form of a particular solution to each ODE. The question asks me to represent all constants in the solution using "P,Q,R,S,T..etc.", in that order.

1. y'''-9y''+14y'=x2
2. y''-9y'+14y=x2e4x

For the first one I wrote:
yp=Px3+Qx2+Rx

Second one:
yp=(Px2+Qx+R)e4x

Neither of the answers are correct, according to the computer. Where did I go wrong?
For Question 1, rewrite your equation as \displaystyle \displaystyle \begin{align*} u'' - 9u' + 14u = x^2 \end{align*} by letting \displaystyle \displaystyle \begin{align*} u = y' \end{align*}. Then to solve for \displaystyle \displaystyle \begin{align*} u \end{align*}, the homogeneous solution has complementary function

\displaystyle \displaystyle \begin{align*} m^2 - 9m + 14 &= 0 \\ (m - 2)(m - 7) &= 0 \\ m = 2 \textrm{ or } m = 7 \end{align*}

So the homogeneous solution is \displaystyle \displaystyle \begin{align*} C_1 \, e^{2x} + C_2 \, e^{7x} \end{align*}.

Now as for the nonhomogeneous solution, use \displaystyle \displaystyle \begin{align*} u = A\,x^2 + B\,x + C \end{align*}, which means \displaystyle \displaystyle \begin{align*} u' = 2A\,x + B \end{align*} and \displaystyle \displaystyle \begin{align*} u'' = 2A \end{align*}. Substituting into the DE gives

\displaystyle \displaystyle \begin{align*} u'' - 9u' + 14u &= x^2 \\ 2A - 9\left( 2A\,x + B \right) + 14 \left( A\, x^2 + B\, x + C \right) &= x^2 \\ 2A - 18A\,x - 9B + 14A\, x^2 + 14B\, x + 14C &= x^2 \\ 14A\, x^2 + \left( 14B - 18A \right) x + 2A - 9B + 14C &= 1x^2 + 0x + 0 \\ 14A = 1, 14B - 18A = 0, 2A - 9B + 14C &= 0 \\ A = \frac{1}{14}, B = \frac{9}{98}, C &= \frac{67}{1372} \end{align*}

Therefore the nonhomogeneous solution is \displaystyle \displaystyle \begin{align*} \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \end{align*}, so putting the homogeneous and nonhomogeneous solutions together gives the solution for \displaystyle \displaystyle \begin{align*} u \end{align*} as

\displaystyle \displaystyle \begin{align*} u &= C_1\, e^{2x} + C_2\, e^{7x} + \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \\ y' &= C_1 \, e^{2x} + C_2\, e^{7x} + \frac{1}{14}x^2 + \frac{9}{98}x + \frac{67}{1372} \\ y &= \frac{C_1}{2} \, e^{2x} + \frac{C_2}{7} \, e^{7x} + \frac{1}{42} x^3 + \frac{9}{196} x^2 + \frac{67}{1372}x + C_3 \end{align*}