# Thread: First-Order Linear Differential Equation

1. ## First-Order Linear Differential Equation

I am suppose to solve y'=5y; y(0)=1

However, I don't understand how to do this as the standard form for this type of equation seems to be: y' + a(t)y = b(t)

What do I use as b(t)?

Everything I've done so far works out to zero. Could someone explain?

2. $\displaystyle y'-5y=0$

which has the characteristic equation $\displaystyle r-5=0$

so the general solution is $\displaystyle C_{1}e^{5t}$

but $\displaystyle y(0)=1=C_{1}$

so the final solution is $\displaystyle e^{5t}$

3. Originally Posted by Quixotic
I am suppose to solve y'=5y; y(0)=1

However, I don't understand how to do this as the standard form for this type of equation seems to be: y' + a(t)y = b(t)

What do I use as b(t)?

Everything I've done so far works out to zero. Could someone explain?

$\displaystyle y' = 5y$

Since this is separable,

$\displaystyle \int \frac{1}{5y} dy = \int 1 dx$

$\displaystyle \frac{log(y)}{5} + C_1 = x + C_2$

$\displaystyle \frac{log(y)}{5} = x + C_3$ Note: $\displaystyle C_3 = {C_2}-{C_1}$ is also a constant

$\displaystyle log(y) = 5x+ C_4$ note: $\displaystyle C_4 = 5 C_3$ is a constant

$\displaystyle \therefore y = e^ {5x+{c_4}}$

$\displaystyle y = e^{5x} \times e^{C_4}$

Thus,

$\displaystyle y= C e^{5x}$.......(I) This is what Random Variable has gotten in the above post.

Note: $\displaystyle C = e^{C_4}$ is also a constant.

now use y(0)=1 [in(I)] to find C