# Finding a solution to this?

1. Sep 25, 2004

### parwana

Find all solutions to

y'= 3y + 15

and solve the initial value problem-
y'=3y+15
y(0)= -1

so would it be y= 3/2x^2 + 15y + C?????

if not please share what would it be!!!!

2. Sep 25, 2004

### mathwonk

just separate variables. i.e. dy/dx = 3y + 15, so

dy = (3y+15)dx
so

dy/(y+15) = dx. now integrate both sides. and keep the constant of integration on one side.

3. Sep 25, 2004

### arildno

Hint:
y'=3(y+5)
Introduce the new function v(x)=y(x)+5, which implies:
v'=3v
Can you take solve it on your own from here?

4. Sep 26, 2004

### parwana

can someone please solve it entirely, i dont get it at all.

5. Sep 26, 2004

### arildno

Since
v'=3v, clearly
v(x)=Ke^{3x} for some K.
Hence, y(x)=Ke^{3x}-5
K can be determined by the initial condition.

6. Sep 26, 2004

### Phymath

$$dy/dx = 3y+15$$
$$1/(3y+15) dy = dx$$
$$\int dy/(3y+15) = \int dx$$

u = 3y + 15 1/3 du = dy

$$1/3 \int du /(u)= x + c_1$$
$$1/3(ln(u)+c_2) = x + c_1$$
$$1/3 (ln (3y + 15) + c_2) = x + c_1$$
$$ln (3y + 15) = 3(x + c_3)$$
$$y + 5 = e^{x + c}$$

$$y = (e^{x + c}) - 5$$

$$-1 = e^{-1 + c} - 5$$
$$-1+5 = 4 = e^{-1}e^{c}$$
$$4e = e^c$$
$$ln(4e) = c$$
$$2 ln(2) + 1 = c$$

$$y = e^{x + 2 ln 2 + 1} - 5$$

Last edited: Sep 26, 2004
7. Sep 26, 2004

### arildno

Phymath:
You've forgotten to exponentiate your "3" in front of (x+c3)..

8. Sep 26, 2004

### parwana

the x value is 0 guys, not -1, but thanks for the help, i appreciate it.

I have another problem that i tried solving in the dy/dx way, but it gets a bit too complex.

Find the solutions of the differential equation

y'-4y=-10sin(2x)

which has the form y= Asin(2x)+Bcos(2x)

9. Sep 26, 2004

### arildno

Just plug in for y' and y; you'll get 2 equations by which you may determine A and B