How to solve a y'=3y+15 initial value problem?

[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!

 

[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.

 

[arildno]

Hint:
Rewrite your differential equation as:
y'=3(y+5)
Introduce the new function v(x)=y(x)+5, which implies:
v'(x)=y'(x), and your differential equation in y now reads in v:
v'=3v
Can you take solve it on your own from here?

 

[parwana]

can someone please solve it entirely, i don't get it at all.

 

[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.

 

[Phymath]

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

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

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

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

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

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

 

[arildno]

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

 

[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)

 

[arildno]

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