# find the general solution of y"-5y'+6y=0

#### horsegirl0121

##### New member
i need help with this differential equation:

find the general solution of this equation
y"-5y'+6y=0

b) find the particular solution for the equation in (a) satisfying the condition that y=1 & dy/dx=-2 when x=0

c) find the general solution of the differential equation
y"-5y'+6y=e^x

thank you

#### soccerball3211

##### Junior Member
I got y=C(sub 1)e^2x+C(sub 2)e^3x for the general solution

Can anyone verify if this is correct?

#### daon

##### Senior Member
This question is perfectly straight forward.

y"-5y'+6y=0

First find the General Solution:
$$\displaystyle \lambda ^2 -5 \lambda + 6 = 0 \,\, \Rightarrow (\lambda -2)(\lambda-3) = 0 \\ \text{So, } \lambda_1 = 2 \text{ and } \lambda_2 = 3 \\ \\$$

Since these zeros of the characteristic equation are different, we use the general form:
$$\displaystyle \\ \\ y(x) = C_1 e^{\lambda_1 x} + C_2 e^{\lambda_2 x} \\ y(x) = C_1 e^{2 x} + C_2 e^{3 x}$$

Now we want a particular solution, so we will use your given IVs that:

1. y(0) = 1
2. y'(0) = -2

So:

$$\displaystyle y'(x) = 2C_1 e^{2x} + 3C_2e^{3x} \\ y'(0) = 2C_1 + 3C_2 = -2 \\ C_1 = \frac{1}{2} (-3C_2-2)$$

Also,

$$\displaystyle y(0) = C_1 + C_2 = 1\\ C_1 = 1-C_2\\ \\ \\$$

Therefore, \\

$$\displaystyle 1-C_2 = \frac{1}{2} (-3C_2-2)\\ 2-2C_2 = -3C_2-2 \\ 2+C_2 = -2 \\ C_2 = -4 \\ \\ So, \,\, C_1 = 5$$

Finally, we have:

$$\displaystyle y(x) = 5e^{2x} - 4e^{3x}$$

Note: Part C is just as easy and is also perfectly straightforward, but is non-homogeneous, so it requires a couple extra steps. Please try this one yourself and post if you have any difficulties.