Solution of the differential equation `(x^(2)+2y^(2))dx-xy dy=0,` when y (9)=0 is

Options

A. `x^(4)=-81(x^(2)+y^(2))`
B. `x^(4)=81(x^(2)+y^(2))`
C. `x^(4)=-9(x^(2)+y^(2))`
D. `x^(4)=9(x^(2)+y^(2))`

Answer

To solve the differential equation \((x^2 + 2y^2)dx - xy dy = 0\) with the initial condition \(y(9) = 0\), we can follow these steps:

### Step 1: Rearranging the Equation
We start with the given differential equation:
\[ (x^2 + 2y^2)dx - xy dy = 0 \]
Rearranging gives:
\[ xy dy = (x^2 + 2y^2)dx \]
Dividing both sides by \(xy\) (assuming \(y \neq 0\)):
\[ \frac{dy}{dx} = \frac{x^2 + 2y^2}{xy} \]

### Step 2: Substituting \(y = vx\)
Let \(y = vx\), where \(v\) is a function of \(x\). Then, we have:
\[ \frac{dy}{dx} = v + x\frac{dv}{dx} \]
Substituting this into our equation gives:
\[ v + x\frac{dv}{dx} = \frac{x^2 + 2(vx)^2}{x(vx)} \]
This simplifies to:
\[ v + x\frac{dv}{dx} = \frac{x^2 + 2v^2x^2}{vx^2} = \frac{1 + 2v^2}{v} \]

### Step 3: Rearranging the Equation
Rearranging gives:
\[ x\frac{dv}{dx} = \frac{1 + 2v^2}{v} - v \]
Combining the terms on the right-hand side:
\[ x\frac{dv}{dx} = \frac{1 + 2v^2 - v^2}{v} = \frac{1 + v^2}{v} \]

### Step 4: Separating Variables
Now we can separate variables:
\[ \frac{v}{1 + v^2} dv = \frac{dx}{x} \]

### Step 5: Integrating Both Sides
Integrating both sides:
\[ \int \frac{v}{1 + v^2} dv = \int \frac{dx}{x} \]
The left side can be integrated using the substitution \(t = 1 + v^2\), giving:
\[ \frac{1}{2} \ln(1 + v^2) = \ln|x| + C \]

### Step 6: Exponentiating Both Sides
Exponentiating both sides gives:
\[ 1 + v^2 = Cx^2 \]
Substituting back \(v = \frac{y}{x}\):
\[ 1 + \left(\frac{y}{x}\right)^2 = Cx^2 \]
Multiplying through by \(x^2\):
\[ x^2 + y^2 = Cx^4 \]

### Step 7: Applying the Initial Condition
Now we apply the initial condition \(y(9) = 0\):
\[ 9^2 + 0^2 = C(9^4) \]
This simplifies to:
\[ 81 = C \cdot 6561 \]
Thus, we find:
\[ C = \frac{81}{6561} = \frac{1}{81} \]

### Final Solution
Substituting \(C\) back into the equation:
\[ x^2 + y^2 = \frac{1}{81}x^4 \]
This can be rewritten as:
\[ 81(x^2 + y^2) = x^4 \]

### Summary
The solution to the differential equation \((x^2 + 2y^2)dx - xy dy = 0\) with the initial condition \(y(9) = 0\) is:
\[ 81(x^2 + y^2) = x^4 \]