Solve the initial value problem: $$ (\ln (y))^{4} \frac{d y}{d x}=x^{4} y, \quad y(1)=e^{2} $$ $$ y(x)= $$

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We start with the differential equation

$$
(\ln(y))^4 \frac{dy}{dx} = x^4 y, \quad y(1)=e^2.
$$

**Step 1. Change of variable**

Let

$$
u = \ln(y).
$$

Then

$$
y = e^u \quad \text{and} \quad \frac{dy}{dx} = e^u \frac{du}{dx}.
$$

**Step 2. Substitute into the original equation**

Substitute the expressions for $$ y $$ and $$ \frac{dy}{dx} $$ into the equation:

$$
(\ln(y))^4 \cdot e^u \frac{du}{dx} = x^4 e^u.
$$

Since $$ \ln(y) = u $$ and $$ e^u $$ is nonzero, we can divide both sides by $$ e^u $$:

$$
u^4 \frac{du}{dx} = x^4.
$$

**Step 3. Separate variables and integrate**

Separate the variables $$ u $$ and $$ x $$:

$$
u^4\, du = x^4\, dx.
$$

Integrate both sides:

$$
\int u^4\, du = \int x^4\, dx.
$$

The integrals are computed as:

$$
\frac{u^5}{5} = \frac{x^5}{5} + C,
$$

or equivalently

$$
u^5 = x^5 + C_1,
$$

where $$C_1 = 5C$$ is an arbitrary constant.

**Step 4. Determine the constant $$C_1$$ using the initial condition**

Recall that $$ u = \ln(y) $$ and from the initial condition $$ y(1)=e^2 $$ we have

$$
u(1) = \ln(e^2) = 2.
$$

Substitute $$ x=1 $$ and $$ u=2 $$ into the equation:

$$
2^5 = 1^5 + C_1.
$$

That is,

$$
32 = 1 + C_1 \quad \Longrightarrow \quad C_1 = 31.
$$

So the equation becomes

$$
u^5 = x^5 + 31.
$$

**Step 5. Solve for $$u$$ and then for $$y$$**

Solve for $$ u $$ by taking the fifth root:

$$
u = \left( x^5 + 31 \right)^{\frac{1}{5}}.
$$

Recall that $$ u = \ln(y) $$, so we have

$$
\ln(y) = \left( x^5 + 31 \right)^{\frac{1}{5}}.
$$

Exponentiate both sides to solve for $$ y $$:

$$
y = e^{\left( x^5 + 31 \right)^{\frac{1}{5}}}.
$$

Thus, the solution to the initial value problem is

$$
y(x)=e^{\left( x^5+31 \right)^{\frac{1}{5}}}.
$$
The solution to the initial value problem is $$ y(x) = e^{\left( x^5 + 31 \right)^{\frac{1}{5}}} $$.

Solve the initial value problem:

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