First-Order Linear Differential Equation

**Quixotic** · May 1st 2010, 10:42 AM

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?

**Random Variable** · May 1st 2010, 10:47 AM

$y'-5y=0$

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

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

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

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

**harish21** · May 1st 2010, 10:58 AM

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?

$y' = 5y$

Since this is separable,

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

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

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

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

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

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

Thus,

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

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

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

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