Consider the differential equation dy/dx=2x/y

Expert intro: This is a standard separable differential equation. Separate the variables, integrate both sides, then use the initial condition to determine the constant and identify the coefficients in the required square-root form.

Detailed walkthrough

We are given

$\frac{dy}{dx}=\frac{2x}{y}$

with initial condition

$f(3)=5.$

Step 1: Separate variables

Multiply both sides by $y\,dx$:

$y\,dy = 2x\,dx$

Step 2: Integrate both sides

$\int y\,dy = \int 2x\,dx$

$\frac{1}{2}y^2 = x^2 + C$

Multiply by 2:

$y^2 = 2x^2 + C'$

So

$y = \sqrt{2x^2 + C'}$

since the initial condition gives a positive value.

Step 3: Use the initial condition

Given $f(3)=5$:

$25 = 2(3)^2 + C'$

$25 = 18 + C'$

$C' = 7$

Thus

$f(x)=\sqrt{2x^2+7}$

So in the form $f(x)=\sqrt{ax^2+b}$, we have

$a=2,\quad b=7.$

Therefore,

$\boxed{a+b=9}$

💡 Pitfall guide

A common mistake is to forget the constant after integrating and jump straight to the square-root form. Another error is choosing the negative root without checking the initial condition. Since $f(3)=5>0$, the positive branch is the correct one.

🔄 Real-world variant

If the initial condition were negative, such as $f(3)=-5$, the algebraic form after separation would still be $y^2=2x^2+7$, but the correct solution branch would be $y=-\sqrt{2x^2+7}$. The value of $a+b$ would still be 9.

🔍 Related terms

separable differential equation, initial condition, integration constant