A function satisfying `int_0^1f(tx)dt=nf(x)`, where `x>0` is

Answer

To solve the problem, we need to find the function \( f(x) \) that satisfies the equation:

\[ \int_0^1 f(tx) \, dt = n f(x) \]

where \( x > 0 \).

### Step-by-Step Solution:

**Step 1: Change of Variables**

We start by making a substitution in the integral. Let \( u = tx \). Then, the differential \( dt \) can be expressed as:

\[ dt = \frac{du}{x} \]

Next, we need to change the limits of integration. When \( t = 0 \), \( u = 0 \). When \( t = 1 \), \( u = x \). Thus, the integral becomes:

\[ \int_0^1 f(tx) \, dt = \int_0^x f(u) \frac{du}{x} \]

This simplifies to:

\[ \frac{1}{x} \int_0^x f(u) \, du \]

**Step 2: Substitute Back into the Equation**

Substituting this back into the original equation gives:

\[ \frac{1}{x} \int_0^x f(u) \, du = n f(x) \]

Multiplying both sides by \( x \) results in:

\[ \int_0^x f(u) \, du = n x f(x) \]

**Step 3: Differentiate Both Sides**

Next, we differentiate both sides with respect to \( x \). Using the Fundamental Theorem of Calculus on the left side, we have:

\[ f(x) = n f(x) + n x f'(x) \]

**Step 4: Rearranging the Equation**

Rearranging the equation gives:

\[ f(x) - n f(x) = n x f'(x) \]

This simplifies to:

\[ (1 - n) f(x) = n x f'(x) \]

**Step 5: Separation of Variables**

Now, we can separate the variables:

\[ \frac{f'(x)}{f(x)} = \frac{1 - n}{n x} \]

**Step 6: Integrate Both Sides**

Integrating both sides, we have:

\[ \int \frac{f'(x)}{f(x)} \, dx = \int \frac{1 - n}{n x} \, dx \]

The left side integrates to \( \ln |f(x)| \), and the right side integrates to \( \frac{1 - n}{n} \ln |x| + C \), where \( C \) is a constant of integration. Thus, we have:

\[ \ln |f(x)| = \frac{1 - n}{n} \ln |x| + C \]

**Step 7: Exponentiating Both Sides**

Exponentiating both sides yields:

\[ f(x) = e^C |x|^{\frac{1 - n}{n}} \]

Let \( K = e^C \), which is a constant. Therefore, we can write:

\[ f(x) = K x^{\frac{1 - n}{n}} \]

### Final Result

Thus, the function \( f(x) \) that satisfies the given integral equation is:

\[ f(x) = K x^{\frac{1 - n}{n}} \]

where \( K \) is a constant.