Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=\frac{4 \sqrt{y} \ln x}{x}\) and \(y=1\) when \(x=e\)

Short answer

The solution to the differential equation is \(y = [2x \ln{x} - 2x + 1 + 2e]^2\), which is valid for all \(x > 0\), and \(y \geq 0\).

Step-by-step solution

Separate the Variables

The given differential equation is \(\frac{d y}{d x}=\frac{4 \sqrt{y} \ln x}{x}\).\n We first rearrange the equation to get all terms involving \(y\) on one side of the equation and those involving \(x\) on the other side.\n Multiply both sides by \(x\) and divide by \(\sqrt{y}\) to get the equation in the separated form: \(\frac{1}{\sqrt{y}} dy = 4 \ln{x} dx \).

Integrate Both Sides

Now, integrate both sides of the equation: \( \int \frac{1}{\sqrt{y}} dy = 4 \int \ln{x} dx \).\n This results to \(2 \sqrt{y} = 4(x \ln{x} - x) + C \), where C is the constant of integration.

Apply the Initial Condition

Given that \(y=1\) when \(x=e\), substitute these values into the equation to find the constant: \(2 \sqrt{1} = 4(e \ln{e} - e) + C\), this simplifies to: \(2 = 4(1 - e) + C\). Solving for \(C\) yields \(C = 2 + 4e\).\n Insert this value of \(C\) back into the equation results to: \(2 \sqrt{y} = 4(x \ln{x} - x) + 2 + 4e\).

Solve for y

Finally, solve for \(y\) by squaring both sides of the equation: \(y = [2x \ln{x} - 2x + 1 + 2e]^2\).\n This is the solution of the differential equation which is valid for all \(x>0\), and \(y \geq 0\).