Ordinary Differential Equations/First Order Linear 4

1)

$$y'+3y=sin(x)\,\!$$

Step 1: Find $$e^{\int P(x)dx}$$

$$\int 3dx=3x+C$$

$$e^{\int P(x)dx}=Ce^{3x}$$

Letting C=1, we get $$e^{3x}$$

Step 2: Multiply through

$$e^{3x}y'+e^{3x}3y=e^{3x}sin(x)\,\!$$

Step 3: Recognize that the left hand is $$\frac{d}{dx} e^{\int P(x)dx}y$$

$$\frac{d}{dx} e^{3x}y=e^{3x}sin(x)$$

Step 4: Integrate

$$\int (\frac{d}{dx} e^{3x}y)dx=\int e^{3x}sin(x)dx$$

$$e^{3x}y=\frac{e^{3x}(3sin(x)-cos(x))}{10}+C$$

Step 5: Solve for y

$$y=\frac{3sin(x)-cos(x)}{10}+\frac{C}{e^{3x}}$$

2)

$$y'+\frac{1}{x+3}y=7x^2+4x$$

Step 1: Find $$e^{\int P(x)dx}$$

$$\int \frac{dx}{x+3}=ln(x+3)+C$$

$$e^{\int P(x)dx}=Cx+3C$$

Letting C=1, we get $$x+3$$

Step 2: Multiply through

$$(x+3)y'+(x+3)y=(x+3)(7x^2+4x)\,\!$$

Step 3: Recognize that the left hand is $$\frac{d}{dx} e^{\int P(x)dx}y$$

$$\frac{d}{dx} (x+3)y=(x+3)(7x^2+4x)$$

Step 4: Integrate

$$\int (\frac{d}{dx} (x+3)y)dx=\int (x+3)(7x^2+4x)dx$$

$$(x+3)y=\frac{7x^4}{4}+\frac{25x^3}{3}+6x^2+C$$

Step 5: Solve for y

$$y=\frac{\frac{7x^4}{4}+\frac{25x^3}{3}+6x^2+C}{x+3}$$

3)

$$(x^4e^x-2mxy^2)dx+2mx^2ydy$$

Step 1: Rearrange

$$2y\dfrac{dy}{dx}-\frac{2y^2}{x}=-x^2e^x$$

Step 2: Substitute $$z=y^2 \implies \dfrac{dz}{dx}=2y\dfrac{dy}{dx}$$

$$\dfrac{dz}{dx}-2\frac{z}x=-x^2e^x$$

Step 3: Find $$e^{\int P(x)dx}$$

Integrating Factor$$=\frac1{x^2}$$

Step 4: Solve for y

$$y(x)=x^2\int \frac{-x^2e^xdx}{mx^2}=x^2\int \frac{-e^x}m dx=\frac{-x^2e^x}m+Cx^2$$