MHB Solving 1st order non-linear ODE

[WMDhamnekar]

I want to solve $\d{y}{x}=\frac{3*(2x-7y)+6}{2*(2x-7y)-3}.$ I don't know its step by step solution. But using some trick of solving ordinary differential equation (which I saw on the Internet), I got the following solution:-

$-\frac{17}{21}*(3x-2y)+ln(119y-34x-48)=C$. Now how to solve this answer for y(x)?

I also want to know its step by step solution.

If any member knows the correct answer, he/she may reply with correct answer. Your Wolfram Alpha DiffEq calculator gives this answer$y(x)=\frac{21}{34}\big(W\big(-e^{(289*x)/147+c_1-1} \big)+1\big)+\frac{4*x-3}{14}$

I don't understand how to verify my answer and (1) are equivalent or not?

 

[MarkFL]

I would try the substitution:

$$u=2x-7y\implies \d{u}{x}=2-7\d{y}{x}\implies \d{y}{x}=\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}$$

Our ODE then becomes:

$$\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}=\frac{3u-6}{2u-3}$$

$$\d{u}{x}=\frac{36-17u}{2u-3}$$

$$\int\frac{2u-3}{36-17u}\,du=\int\,dx$$

$$-34u-21\ln(17u-36)+72=289x+C$$

Back-substitute for \(u\):

$$-34(2x-7y)-21\ln(17(2x-7y)-36)+72=289x+C$$

I thus get the implicit solution:

$$\frac{17}{3}(2y-3x)-\ln(17(2x-7y)-36)=c_1$$

 

[topsquark]

I would try the substitution:

$$u=2x-7y\implies \d{u}{x}=2-7\d{y}{x}\implies \d{y}{x}=\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}$$

Our ODE then becomes:

$$\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}=\frac{3u-6}{2u-3}$$

$$\d{u}{x}=\frac{36-17u}{2u-3}$$

$$\int\frac{2u-3}{36-17u}\,du=\int\,dx$$

$$-34u-21\ln(17u-36)+72=289x+C$$

Back-substitute for \(u\):

$$-34(2x-7y)-21\ln(17(2x-7y)-36)+72=289x+C$$

I thus get the implicit solution:

$$\frac{17}{3}(2y-3x)-\ln(17(2x-7y)-36)=c_1$$

Nice! (Bow)

-Dan

 

[WMDhamnekar]

I would try the substitution:

$$u=2x-7y\implies \d{u}{x}=2-7\d{y}{x}\implies \d{y}{x}=\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}$$

Our ODE then becomes:

$$\frac{2}{7}-\frac{1}{7}\cdot\d{u}{x}=\frac{3u-6}{2u-3}$$

$$\d{u}{x}=\frac{36-17u}{2u-3}$$

$$\int\frac{2u-3}{36-17u}\,du=\int\,dx$$

$$-34u-21\ln(17u-36)+72=289x+C$$

Back-substitute for \(u\):

$$-34(2x-7y)-21\ln(17(2x-7y)-36)+72=289x+C$$

I thus get the implicit solution:

$$\frac{17}{3}(2y-3x)-\ln(17(2x-7y)-36)=c_1$$

Hello,
Taking clue from your answer,
Let u=2x-7y, $\Rightarrow\d{u}{x}=2-7\d{y}{x}\Rightarrow\d{y}{x}=\frac27-\frac17\d{u}{x}$

Our ODE then becomes $\frac27-\frac17\d{u}{x}=\frac{3u+6}{2u-3}\Rightarrow \d{u}{x}=\frac{-17u-48}{2u-3}$

$\displaystyle\int \frac{2u-3}{-17u-48}=\displaystyle\int dx$

$-34u+147*\ln{|17u+48|}=289x+C$

$-34*(2x-7y)+147*\ln{|17(2x-7y)+48|}=289x+C$$-357x+238y+147*\ln{|34x-119y+48|}=C$ is the final answer.

So It proves that there is no difference between the trick used answer and my computed step by step answer.

What is the meaning of W in the answer given by WolframAlpha DiffEq calculator?

 

[HallsofIvy]

I can't be certain but I suspect "W" is "Lambert's W function", also called the "omega function" or "product logarithm". It is defined as the inverse function to f(x)= xe^x. https://en.wikipedia.org/wiki/Lambert_W_function.