Differential Equations/First Order Linear 4

Current revision (unreviewed)

1)

$y' + 3 y = s i n (x)$

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

$\int 3dx=3x+C$

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

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

Step 2: Multiply through

$e 3 x y' + e 3 x 3 y = e 3 x s i n (x)$

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

$(x + 3) y' + (x + 3) y = (x + 3)(7 x 2 + 4 x)$

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

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