Question

Solve dy/DX +xy= y^2e^x^2/2sin x

Answer 1

Answer on Question #66141 – Math – Differential Equations

Question

Solve

$$\frac{dy}{dx} + xy = \frac{y^2 e^{x^2}}{2 \sin x}$$

Solution

$$\frac{dy}{dx} + xy = \frac{y^2 e^{x^2}}{2 \sin x'}$$ $$\frac{y'}{y^2} + \frac{x}{y} = \frac{e^{x^2}}{2 \sin x}$$

is the first-order nonlinear ordinary differential equation

First we solve the linear differential equation in order to use the method of variation of the constant

$$\frac{y'}{y^2} + \frac{x}{y} = 0$$ $$\frac{y'}{y} + x = 0$$ $$\frac{dy}{y} + xdx = 0$$ $$\int \frac{dy}{y} + \int xdx = C$$ $$\ln(y) = -\frac{x^2}{2} + C$$ $$y = A e^{\frac{-x^2}{2}},$$

$C$ and $A$ are arbitrary constants of integration

$$y = A e^{\frac{-x^2}{2}}$$

We use the method of variation of constant. Let $A$ be a function which depends on $x$:

$$A = A(x)$$

We find the derivative and substitute $y$ and $y'$ in the original equation:

$$y' = A' e^{\frac{-x^2}{2}} + A(-x) e^{\frac{-x^2}{2}}$$ $$A' e^{\frac{-x^2}{2}} + A(-x) e^{\frac{-x^2}{2}} + A x e^{\frac{-x^2}{2}} = \frac{e^{x^2}}{2 \sin x} (A e^{\frac{-x^2}{2}})^2$$ $$A ^ {\prime} e ^ {- \frac {x ^ {2}}{2}} = \frac {e ^ {x ^ {2}}}{2 \sin x} A ^ {2} e ^ {- x ^ {2}}$$

Solving the equation

$$\frac {d A}{A ^ {2}} = \frac {e ^ {\frac {x ^ {2}}{2}} d x}{2 \sin x}$$ $$\frac {- 1}{A} = \int \frac {e ^ {\frac {x ^ {2}}{2}} d x}{2 \sin x} + B,$$

$B$ is an integration constant.

The integral is not expressed in elementary functions. Therefore we introduce

$$I (x) = \int \frac {e ^ {\frac {x ^ {2}}{2}} d x}{2 \sin x}.$$

Finally we get

$$\frac {- 1}{A} = I (x) + B,$$ $$A = \frac {- 1}{I (x) + B},$$ $$\frac {y (x)}{e ^ {- \frac {x ^ {2}}{2}}} = - \frac {1}{I (x) + B},$$ $$y (x) = \frac {- e ^ {\frac {- x ^ {2}}{2}}}{I (x) + B}.$$

Answer: $y(x) = \frac{-e^{-\frac{x^2}{2}}}{I(x) + B}$ .

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