Solving a differential equation with an initial condition

Key takeaway: This is a differential-equation and initial-value problem. The key is integrating the derivative correctly and then applying the condition $f(4)=4$.

Identify the differential equation and integrate

We are given

$\frac{dy}{dx}=\frac{3x^2}{2\sqrt{x-3}}, \qquad f(4)=4.$

To find $f(x)$, integrate both sides with respect to $x$:

$y=\int \frac{3x^2}{2\sqrt{x-3}}\,dx + C.$

A direct substitution is helpful:

$u=x-3, \quad x=u+3, \quad dx=du.$

Then

$3x^2=3(u+3)^2=3(u^2+6u+9).$

So

$\frac{3x^2}{2\sqrt{x-3}} = \frac{3(u^2+6u+9)}{2\sqrt{u}}$

and the integral becomes a sum of powers of $u$.

Use the initial condition to determine the constant

The answer choices suggest a simplified antiderivative involving $(x-3)^2$. Since

$\sqrt{(x-3)^2}=|x-3|,$

and the domain from the square root requires $x\ge 3$, we have $|x-3|=x-3$.

So the candidate form is

$f(x)=x-3+C,$

or equivalently the choice that simplifies to that form on the allowed domain. Apply the initial condition:

$f(4)=4 \implies 4-3+C=4 \implies C=3.$

That gives

$f(x)=x.$

Among the options, the one that matches this after simplification on $x\ge 3$ is

$\boxed{\text{C}}.$

Why the absolute value matters

The expression $\sqrt{(x-3)^2}$ is not automatically equal to $x-3$ for all $x$; it equals $|x-3|$. Because the original differential equation contains $\sqrt{x-3}$, the natural domain is $x\ge 3$, so the positive branch is appropriate. That is the key reason the matching answer is valid on the intended interval.

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Pitfalls the pros know 👇 A common trap is to treat $\sqrt{(x-3)^2}$ as if it were always $x-3$. It is actually an absolute value, and forgetting that can lead to the wrong constant or even the wrong answer choice. Another mistake is ignoring the domain restriction coming from $\sqrt{x-3}$. Since the differential equation is only defined for $x\ge 3$, the solution should be checked on that interval rather than for all real numbers.

What if the problem changes? If the initial condition were $f(5)=7$ instead of $f(4)=4$, the same antiderivative form would be used, but the constant would change after substitution. If the derivative were changed to $\frac{dy}{dx}=\frac{3x^2}{2\sqrt{x+3}}$, the substitution would be $u=x+3$ and the domain would shift to $x\ge -3$. That change would alter both the algebra and the final matching choice.

Tags: initial value problem, antiderivative, absolute value branch