# IVP find constants to solution

#### find_the_fun

##### Active member
The given family of functions is the general solution of the D.E. on the indicated interval. Find a member of the family that is a solution of the initial-value problem.

$$\displaystyle y=c_1x+c_2x\ln{x}$$ on $$\displaystyle (0, \infty)$$ and $$\displaystyle x^2y''-xy'+y=0$$ and y(1)=3, y'(1)=-1

So plugging in y(1)=3 gives $$\displaystyle 3=c_1+c_2\ln{1}$$ and then take the derivative to get $$\displaystyle y'=c_1+c_2 \ln{x} +c_2$$ subbing in $$\displaystyle -1=C-1+c_2\ln{1}+c_2$$

adding 3 times the second equation to the first give $$\displaystyle 0=4c_1+4c_2\ln{1}+3c_2$$
What next?

#### MarkFL

Staff member
Okay, we have:

$$\displaystyle y(1)=c_1=3$$ (recall $\log_a(1)=0$)

$$\displaystyle y'(1)=c_1+c_2=-1$$

Now the system is easier to solve.

#### find_the_fun

##### Active member
Okay, we have:

$$\displaystyle y(1)=c_1=3$$ (recall $\log_a(1)=0$)

$$\displaystyle y'(1)=c_1+c_2=-1$$

Now the system is easier to solve.
Ok but we never used $$\displaystyle x^2y''-xy'+y=0$$ Why was that in the question?

#### MarkFL

Ok but we never used $$\displaystyle x^2y''-xy'+y=0$$ Why was that in the question?
$$\displaystyle x=e^t$$
$$\displaystyle y=x^r$$