The solution curve, of the differential equation passing through the point (0,1) is a conic, whose vertex lies on the line:

Engineering

Mathematics

Variable Separable Differential Equation

Standard Parabola

Question

The solution curve, of the differential equation $2 y \frac{dy}{dx} + 3 = 5 \frac{dy}{dx}$ passing through the point (0,1) is a conic, whose vertex lies on the line:

2x + 3y = –9

2x + 3y = 6

2x + 3y = 9

2x + 3y = –6

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2ydy + 3dx = 5dy

⇒ y² + 3x = 5y + c

passes through point (0, 1)

1 = 5 + c ⇒ c = –4

y² + 3x = 5y – 4

y² – 5y = –3x – 4

⇒ ${(y - \frac{5}{2})}^{2} = - 3 x - 4 + \frac{25}{4}$

⇒ ${(y - \frac{5}{2})}^{2} = - 3 (x - \frac{3}{4})$

so vertex = $(\frac{3}{4}, \frac{5}{2})$

It lies on the line 2x + 3y = 9