Question

Obtain the general solution of the following differential equations.

1. 𝑑𝑦 + 𝑦𝑒^x𝑑𝑥 = 𝑒^x𝑑𝑥

2. 𝑥𝑑𝑦 = (𝑒^x − 2𝑦)𝑑𝑥

Obtain the particular solution satisfying the indicated conditions.

3. (2𝑦 − 𝑥³)𝑑𝑥 = 𝑥𝑑𝑦, 𝑦(2) = 4

Answer 1

1.

$$dy=e^x(1-y)dx$$

$$\dfrac{dy}{1-y}=e^xdx$$

Integrate

$$\int \dfrac{dy}{1-y}=\int e^xdx$$

$$-\ln|1-y|=e^x-\ln C$$

$$1-y=Ce^{-e^x}$$

$$y=1-Ce^{-e^x}$$

2.

$$𝑥𝑑𝑦 = (𝑒^x − 2𝑦)𝑑𝑥$$

$$y'+2\dfrac{y}{x}=\dfrac{e^x}{x}$$

Integrating factor

$$\mu(x)=x^2$$

$$x^2y'+2xy=xe^x$$

$$d(x^2y)=xe^xdx$$

Integrate

$$\int d(x^2y)=\int xe^xdx$$

$$\int xe^xdx$$

$$\int udv=uv-\int vdu$$

$$u=x, du=dx$$

$$dv=e^xdx, v=\int e^x dx=e^x$$

$$\int xe^xdx=xe^x-\int e^xdx=xe^x-e^x+C$$

Then

$$x^2y=xe^x-e^x+C$$

$$y=\dfrac{e^x}{x}-\dfrac{e^x}{x^2}+\dfrac{C}{x^2}$$

3.

$$y'-\dfrac{2}{x}y=-x^2$$

Integrating factor

$$\mu(x)=\dfrac{1}{x^2}$$

$$\dfrac{1}{x^2}y'-\dfrac{2}{x^3}y=-1$$

$$d(\dfrac{y}{x^2})=-dx$$

Integrate

$$\int d(\dfrac{y}{x^2})=-\int dx$$

$$\dfrac{y}{x^2}=-x+C$$

$$y=-x^3+Cx^2$$

$𝑦(2) = 4$

$$4=-(2)^3+C(2)^2=>C=3$$

The particular solution satisfying the indicated conditions is

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