Find the function $$ y=y(x)( $$ for $$ x0) $$ which satisfies the separable differential equation $$ \frac{d y}{d x}=\frac{9+20 x}{x y^{2}} ; x0 $$ with the initial condition $$ y(1)=3 $$ $$ y= $$

Question

Find the function $$ y=y(x)( $$ for $$ x>0) $$ which satisfies the separable differential equation $$ \frac{d y}{d x}=\frac{9+20 x}{x y^{2}} ; x>0 $$ with the initial condition $$ y(1)=3 $$ $$ y= $$

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We start with the differential equation
$$
\frac{dy}{dx} = \frac{9 + 20x}{x\, y^2}\quad (x>0)
$$
and the initial condition
$$
y(1) = 3.
$$

**Step 1. Separate the variables.**

Multiply both sides by $$ y^2 $$ and by $$ dx $$ to obtain:
$$
y^2\, dy = \frac{9 + 20x}{x}\, dx.
$$
We can write the right-hand side as:
$$
\frac{9 + 20x}{x} = \frac{9}{x} + 20.
$$

**Step 2. Integrate both sides.**

Integrate the left-hand side with respect to $$ y $$ and the right-hand side with respect to $$ x $$:
$$
\int y^2\, dy = \int \left(\frac{9}{x} + 20\right) dx.
$$

The integrals evaluate as:
$$
\int y^2\, dy = \frac{y^3}{3}+C_1,\quad \int \left(\frac{9}{x} + 20\right) dx = 9\ln|x| + 20x+C_2.
$$
We can combine the constants $$ C_1 $$ and $$ C_2 $$ into a single constant $$ C $$, so that:
$$
\frac{y^3}{3} = 9\ln|x| + 20x + C.
$$
Since $$ x>0 $$, we have $$ \ln|x|=\ln x $$.

**Step 3. Use the initial condition to find $$ C $$.**

Plug in $$ x=1 $$ and $$ y=3 $$ into the equation:
$$
\frac{3^3}{3} = 9\ln 1 + 20(1) + C.
$$
Since $$3^3 = 27$$ and $$ \ln 1 = 0 $$, this becomes:
$$
\frac{27}{3} = 0 + 20 + C \quad \Longrightarrow \quad 9 = 20 + C.
$$
Thus,
$$
C = 9 - 20 = -11.
$$

**Step 4. Write the final solution.**

Substitute $$ C = -11 $$ back into the integrated equation:
$$
\frac{y^3}{3} = 9\ln x + 20x - 11.
$$

Multiplying through by 3 gives:
$$
y^3 = 27\ln x + 60x - 33.
$$

Taking the cube root on both sides, we obtain:
$$
y = \sqrt[3]{27\ln x + 60x - 33}.
$$

Thus, the solution of the differential equation satisfying the initial condition is:
$$
y = \sqrt[3]{27\ln x + 60x - 33}\quad (x>0).
$$
The solution is $$ y = \sqrt[3]{27\ln x + 60x - 33} $$ for $$ x > 0 $$.

Find the function for which satisfies the separable differential equation

with the initial condition

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Find the function for which satisfies the separable differential equation with the initial condition