IVP find constants to solution

[find_the_fun]

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]

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. :D

 

[find_the_fun]

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. :D

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?

That was to show you which ODE the given solution satisfies. You are right though, we didn't need it to find the particular solution satisfying the given conditions. The given ODE is a Cauchy-Euler equation which may be solved either by making the substitution:

\(\displaystyle x=e^t\)

which will given you a homogeneous equation with constant coefficients, or by assuming a solution of the form:

\(\displaystyle y=x^r\)

which will give you an indicial equation with a repeated root to which you may apply the method of reduction of order.

 

[blue_raver22]

To find the constants c_1 and c_2, we can solve the system of equations formed by equating the coefficients of c_1 and c_2 in the two equations we have. This will give us two equations with two unknowns, which we can solve using standard algebraic techniques.

First, we can simplify the second equation by substituting ln{1} with 0, since ln{1}=0. This gives us -1=c_1+c_2. We can then substitute this value of c_1 into the first equation to get 3=c_1+c_2\ln{1}. Since ln{1}=0, this simplifies to 3=c_2. So we have found the values of both c_1 and c_2, which are c_1=4 and c_2=3.

Therefore, a member of the given family of functions that satisfies the initial-value problem is y=4x+3x\ln{x}. This solution can be verified by plugging it into the differential equation and checking that it satisfies the initial conditions y(1)=3 and y'(1)=-1.