sibuzwaW

2021-09-26

Solve differential equation.

$\frac{dy}{dx}=(x+y-1)+\frac{x+y}{\mathrm{log}(x+y)}$

tabuordy

Skilled2021-09-27Added 90 answers

Step 1 Given

$\frac{dy}{dx}=(x+y-1)+\frac{x+y}{\mathrm{log}(x+y)}$

Step 2 solving Differential Equations

Let x+y=v.

Then,$1+\frac{dy}{dx}+\frac{dv}{dx}$

The given differential equation reduces to

$\frac{dv}{dx}-1=v+\frac{v}{\mathrm{log}v}$

$\frac{dv}{dx}=v+\frac{v}{\mathrm{log}v}$

$\frac{\mathrm{log}v}{v(1+\mathrm{log}v)}dv=dx$

On integrating, we get

$\int \frac{\mathrm{log}v}{(1+\mathrm{log}v)}\frac{1}{v}dv=\int 1.dx+C$

$\int \frac{t-1}{t}dt=x+C$

Where$t=1+\mathrm{log}v$

$t-\mathrm{log}t=x+C$

$(1+\mathrm{log}v)-\mathrm{log}[1+\mathrm{log}v]=x+C$

Step 2 solving Differential Equations

Let x+y=v.

Then,

The given differential equation reduces to

On integrating, we get

Where

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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