MHB Need help setting up Differential Equation to solve

[Logan Land]

dy/dx = (2y-x+5)/(2x-y-4)
y(1)=1

 

[MarkFL]

If we express the given ODE in differential form, we obtain:

$$(2y-x+5)\,dx+(-2x+y+4)\,dy=0$$

It is easy to see by inspection that this is not an exact equation. However, obtaining a special integrating factor seems to be impossible, at least by the technique normally given in an elementary course in ODEs.

So, allow me to ask that you are certain the problem has been copied correctly. :D

 

[Logan Land]

If we express the given ODE in differential form, we obtain:

$$(2y-x+5)\,dx+(-2x+y+4)\,dy=0$$

It is easy to see by inspection that this is not an exact equation. However, obtaining a special integrating factor seems to be impossible, at least by the technique normally given in an elementary course in ODEs.

So, allow me to ask that you are certain the problem has been copied correctly. :D

yes I copied it correctly.
Here is an attachment of the sample exam we were given for our test on Thursday
View attachment 3939

 

[MarkFL]

Okay, let's approach it in the following manner, returning to the original form:

$$\d{y}{x}=\frac{2y-x+5}{2x-y-4}$$

Now, let's let:

$$X=x+a$$

$$Y=y+b$$

and we now have:

$$\d{Y}{X}=\frac{2(Y-b)-(X-a)+5}{2(X-a)-(Y-b)-4}=\frac{2Y-2b-X+a+5}{2X-2a-Y+b-4}$$

And so we find we desire:

$$-2b+a+5=0$$

$$-2a+b-4=0$$

You should find:

$$(a,b)=(-1,2)$$

And this leaves us with:

$$\d{Y}{X}=\frac{2Y-X}{2X-Y}$$

Can you proceed?

 

[MarkFL]

Since several days have passed with no further feedback from the OP, I will finish the problem...

On the right, let's divide each term in the numerator and denominator by $X$ to obtain:

$$\d{Y}{X}=\frac{2\frac{Y}{X}-1}{2-\frac{Y}{X}}$$

Next, we may use the substitution:

$$Y=uX\,\therefore\,\d{Y}{X}=u+X\d{u}{X}$$

And we now have:

$$u+X\d{u}{X}=\frac{2u-1}{2-u}$$

Subtract through by $u$:

$$X\d{u}{X}=\frac{2u-1}{2-u}-u=\frac{2u-1-u(2-u)}{2-u}=\frac{u^2-1}{2-u}$$

Separate variables and use partial fraction decomposition:

$$\left(\frac{1}{u-1}-\frac{3}{u+1}\right)\,du=\frac{2}{X}\,dx$$

Integrating, we obtain:

$$\ln\left|\frac{u-1}{(u+1)^3}\right|=\ln\left|c_1X^2\right|$$

And this implies:

$$\frac{u-1}{(u+1)^3}=c_1X^2$$

Back-substituting for $u$, we have:

$$\frac{\frac{Y}{X}-1}{\left(\frac{Y}{X}+1\right)^3}=c_1X^2$$

$$\frac{X-Y}{\left(X+Y\right)^3}=c_2$$

Back-substitute for $X$ and $Y$:

$$\frac{(x-1)-(y+2)}{\left((x-1)+(y+2)\right)^3}=c_2$$

$$\frac{x-y-3}{\left(x+y+1\right)^3}=c_2$$

Now, use the given initial conditions to find the value of the parameter $c_2$:

$$\frac{1-1-3}{\left(1+1+1\right)^3}=c_2\implies c_2=-\frac{1}{9}$$

Thus, the implicit solution satisfying the given IVP is:

$$\frac{y-x+3}{\left(y+x+1\right)^3}=\frac{1}{9}$$