Thingsto Do's question at Yahoo Answers regarding a first order homogeneous IVP

[MarkFL]

Here is the question:

Solve the following Initial Value Problem Help?

Solve the following Initial Value Problem Help #2

View attachment 1326

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Thingsto Do,

I would first write the ODE in the form $$\frac{dy}{dx}=f(x,y)$$:

$$\frac{dy}{dx}=\frac{104xy-y^2}{x^2}=104\frac{y}{x}-\left(\frac{y}{x} \right)^2$$

Using the substitution $$u=\frac{y}{x}\,\therefore\,\frac{dy}{dx}=u+x\frac{du}{dx}$$ we may write:

$$u+x\frac{du}{dx}=104u-u^2$$

$$x\frac{du}{dx}=103u-u^2$$

Separating variables, we have:

$$\frac{1}{103u-u^2}\,du=\frac{1}{x}\,dx$$

Using the Heaviside cover-up method on the left side, we may obtain the partial fraction decomposition. Factoring the denominator, we may then assume it takes the following form:

$$\frac{1}{u(103-u)}=\frac{A}{u}+\frac{B}{103-u}$$

Covering up the factor $u$ in the denominator of the left side, and evaluating what is left for $u=0$ we obtain:

$$A=\frac{1}{103}$$

Covering up the factor $103-u$ in the denominator of the left side, and evaluating what is left for $u=103$ we obtain:

$$B=\frac{1}{103}$$

Hence:

$$\frac{1}{u(103-u)}=\frac{1}{103}\left(\frac{1}{u}+\frac{1}{103-u} \right)$$

And so the ODE becomes:

$$\frac{1}{103}\left(\frac{1}{u}-\frac{1}{u-103} \right)\,du=\frac{1}{x}\,dx$$

Integrating, we have:

$$\frac{1}{103}\int\left(\frac{1}{u}-\frac{1}{u-103} \right)\,du=\int\frac{1}{x}\,dx$$

$$\ln\left|\frac{u}{u-103} \right|=103\ln\left|Cx \right|=\ln\left|Cx^{103} \right|$$

$$\frac{u}{u-103}=Cx^{103}$$

Solve for $u$:

$$u=\frac{103Cx^{103}}{Cx^{103}-1}$$

Back-substitute for $u$:

$$\frac{y}{x}=\frac{103Cx^{103}}{Cx^{103}-1}$$

Hence:

$$y(x)=\frac{103Cx^{104}}{Cx^{103}-1}$$

Use the initial values to determine the parameter $C$:

$$y(1)=\frac{103C}{C-1}=2\,\therefore\,C=-\frac{2}{101}$$

Thus, the solution satisfying the IVP is:

$$y(x)=\frac{103\left(-\frac{2}{101} \right)x^{104}}{\left(-\frac{2}{101} \right)x^{103}-1}=\frac{206x^{104}}{2x^{103}+101}$$