A curve passing through `(2,3)` and satisfying the differential equation `int_0^x ty(t)dt=x^2y(x),(x >0)` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )+( h ) y^(( i )2( j ))( k )=13 (l)` (m) (b) `( n ) (o) (p) y^(( q )2( r ))( s )=( t )9/( u )2( v ) (w) x (x)` (y) (c) `( d ) (e) (f)(( g ) (h) x^(( i )2( j ))( k ))/( l )8( m ) (n)+( o )(( p ) (q) y^(( r )2( s ))( t ))/( u )(( v ) 18)( w ) (x)=1( y )` (z) (d) `( a a ) (bb) x y=6( c c )` (dd)

Options

A. `x^(2)+y^(2)=13`
B. `y^(2)=9/2x`
C. `x^(2)/8+y^(2)/18=1`
D. `xy=6`

Explanation

To solve the problem, we need to find the equation of a curve that passes through the point (2, 3) and satisfies the given differential equation. Let's break down the solution step by step.

### Step 1: Write down the given differential equation
The differential equation is given as:
\[ \int_0^x t y(t) \, dt = x^2 y(x) \quad (x > 0) \]

### Step 2: Differentiate both sides with respect to \( x \)
Using the Leibniz rule for differentiation under the integral sign, we differentiate the left side:
\[ \frac{d}{dx} \left( \int_0^x t y(t) \, dt \right) = x y(x) \]
The right side differentiates to:
\[ \frac{d}{dx} (x^2 y(x)) = 2x y(x) + x^2 \frac{dy}{dx} \]
Setting both sides equal gives:
\[ x y(x) = 2x y(x) + x^2 \frac{dy}{dx} \]

### Step 3: Rearranging the equation
Rearranging the equation:
\[ x y(x) - 2x y(x) = x^2 \frac{dy}{dx} \]
This simplifies to:
\[ -x y(x) = x^2 \frac{dy}{dx} \]
Dividing both sides by \( x \) (assuming \( x > 0 \)):
\[ -y(x) = x \frac{dy}{dx} \]

### Step 4: Separate variables
Rearranging gives:
\[ \frac{dy}{y} = -\frac{dx}{x} \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \int \frac{dy}{y} = -\int \frac{dx}{x} \]
This results in:
\[ \ln |y| = -\ln |x| + C \]
where \( C \) is the constant of integration.

### Step 6: Exponentiate to solve for \( y \)
Exponentiating both sides gives:
\[ |y| = e^{- \ln |x| + C} = \frac{e^C}{x} \]
Let \( e^C = k \) (a positive constant), so:
\[ y = \frac{k}{x} \]

### Step 7: Find the constant using the point (2, 3)
We know the curve passes through the point (2, 3):
\[ 3 = \frac{k}{2} \implies k = 6 \]

### Step 8: Write the final equation of the curve
Substituting \( k \) back into the equation gives:
\[ y = \frac{6}{x} \]
or equivalently:
\[ xy = 6 \]

### Conclusion
The equation of the curve that passes through (2, 3) and satisfies the differential equation is:
\[ xy = 6 \]