Question

6ylnx+xy'-6ylny=7y y(1)=e^-3

Answer 1

$\displaystyle 6ylnx+xy'-6ylny=7y\\-6yln\frac{y}{x}+xy'=7y\\y=tx;y'=t'x+t\\t'x=6t+6tlnt\\6\int\frac{dx}{x}=\int\frac{d(1+lnt)}{1+lnt}\\ln|1+lnt|=6ln|x|+C\\1+lny-lnx=Cx^6\\1-3=C\to C=-2\\1+ln\frac{y}{x}+2x^6=0$

Answer 2

This is a first-order linear ordinary differential equation. To solve it, we'll use the method of separation of variables and then apply the initial condition to find the particular solution.

The given differential equation is:

6xy' + 6ylny = 7y

Let's rearrange it a bit:

6xy' = 7y - 6ylny

Now, divide both sides by 6x:

y' = (7y - 6ylny) / (6x)

This equation can be separated into variables y and x:

(1 / (7y - 6ylny)) dy = (1 / 6x) dx

Now, integrate both sides:

∫ (1 / (7y - 6ylny)) dy = ∫ (1 / 6x) dx

The left-hand side requires integration by substitution. Let u = 7y - 6lny, then du = (7 - 6/y) dy:

∫ (1 / u) du = ∫ (1 / 6x) dx

ln|u| = (1/6)ln|x| + C₁

Substitute back for u:

ln|7y - 6lny| = (1/6)ln|x| + C₁

Now, let's solve for y:

7y - 6lny = e^((1/6)ln|x| + C₁)

7y - 6lny = C₂ * e^(ln|x|/6)

7y - 6lny = C₂ * (x^(1/6))

Solve for y:

7y - 6lny = C₂ * x^(1/6)

Now, apply the initial condition y(1) = e^(-3):

7(e^(-3)) - 6ln(e^(-3)) = C₂ * (1^(1/6))

7e^(-3) + 18 = C₂

So, the particular solution to the given differential equation with the initial condition is:

7y - 6lny = (7e^(-3) + 18) * x^(1/6)