Second order differential equation Is this correct
Is this solution in this pic correct or not?
I want to solve this equation and that is my understanding.

Continuing from my comment, if there is no typo and the equation is really as you have written it, then the right approach would be to assume $z=y'$.Then we can write $y''-6y'+13=0$ as $z'-6z+13=0$.
Therefore, we solve the above equation as follows:
$$z'= 6z-13$$
$$6z'= 6(6z-13)$$
$$\frac{d(6z-13)}{(6z-13)}=6 \,\ dt$$
$$\int \frac{d(6z-13)}{(6z-13)}=6 \int dt$$
$$\ln |6z-13| =6t+k$$
$$6z-13=e^{6t+k}=c_1e^{6t}$$
$$6\frac{dy}{dt}=c_1e^{6t}+13$$
$$6 \int dy=c_1e^{6t} \int dt+13 \int dt$$
$$y(t)=\frac{c_1e^{6t}}{36}+\frac{13t}{6}+c_2$$
This is the required solution.
And if the equation you have written has some typo error i.e. there is a term of $y$ with coefficient $13$ and no constant term, then your solution is right.
