Question

Solve the following ordinary differential equations: I) ( 2y + X^2 + 1) dy /dx + 2 xy - 9x^2 = 0? 2) d^2y /dx^2 + 3*dy/dx + 2y = X^2

Answer 1

Solve the following ordinary differential equations:

1)

$$(2y + x^2 + 1) \frac{dy}{dx} + 2xy - 9x^2 = 0;$$

2)

$$\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = x^2.$$

**Solution:**

1)

Denote

$$M(x,y) = 2xy - 9x^2, \quad N(x,y) = 2y + x^2 + 1.$$

Because

$$\frac{\partial M}{\partial y} = 2x = \frac{\partial N}{\partial x}$$

then the differential equation

$$(2y + x^2 + 1) \frac{dy}{dx} + 2xy - 9x^2 = 0$$

has general solution

$$U(x,y) = c = \text{const}$$

where

$$\frac{\partial U}{\partial x} = M(x,y) = 2xy - 9x^2,$$ $$\frac{\partial U}{\partial y} = N(x,y) = 2y + x^2 + 1.$$

From first equation we have

$$U(x,y) = \int M(x,y)dx = \int (2xy - 9x^2)dx = yx^2 - 3x^3 + \varphi(y).$$

Thus

$$\frac{\partial}{\partial y} \left(yx^2 - 3x^3 + \varphi(y)\right) = N(x,y) = 2y + x^2 + 1,$$ $$x ^ {2} + \frac {d \varphi}{d y} = 2 y + x ^ {2} + 1,$$ $$\frac {d \varphi}{d y} = 2 y + 1,$$ $$d \varphi = (2 y + 1) d y,$$ $$\int d \varphi = \int (2 y + 1) d y,$$ $$\varphi (y) = y ^ {2} + y.$$

Thus the general solution is

$$y x ^ {2} - 3 x ^ {3} + y ^ {2} + y - c = 0$$

2)

The characteristic equation is

$$k ^ {2} + 3 k + 2 = 0,$$ $$k _ {1} = - 1, k _ {2} = - 2.$$

So general solution of the equation is

$$y (x) = c _ {1} e ^ {- x} + c _ {2} e ^ {- 2 x} + u (x),$$

where

$$u (x) = a x ^ {2} + b x + c, \quad c _ {1} = \text{const}, \quad c _ {2} = \text{const}, \quad a = \text{const}, \quad b = \text{const}, \quad c = \text{const}.$$

Substitution $u(x)$ into the equation we have

$$u ^ {\prime \prime} (x) + 3 u ^ {\prime} (x) + 2 u (x) = x ^ {2},$$ $$2 a + 3 (2 a x + b) + 2 \left(a x ^ {2} + b x + c\right) = x ^ {2},$$ $$2 a x ^ {2} + (6 a + 2 b) x + 2 a + 3 b + 2 c = x ^ {2}.$$

Thus

$$\left\{ \begin{array}{c} 2 a = 1, \\ 6 a + 2 b = 0, \\ 2 a + 3 b + 2 c = 0, \end{array} \right. \to \left\{ \begin{array}{l} a = \frac {1}{2}, \\ b = - 1 \frac {1}{2}, \\ c = 1 \frac {3}{4}. \end{array} \right.$$

So

$$u (x) = \frac {1}{2} x ^ {2} - 1 \frac {1}{2} x + 1 \frac {3}{4}$$

and finally

$$y(x) = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{2} x^2 - 1 \frac{1}{2} x + 1 \frac{3}{4}$$

Answers:

1)

$$y x^2 - 3 x^3 + y^2 + y - c = 0$$

2)

$$y(x) = c_1 e^{-x} + c_2 e^{-2x} + \frac{1}{2} x^2 - 1 \frac{1}{2} x + 1 \frac{3}{4}$$

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