Families of Solutions (Level 3) In this video we are going to continue verifying
general solutions of differential equations. This time around we will verify implicit family
of solutions. Let’s go ahead and try the first example.
Verify that the indicated general solution satisfies the differential equation.
The differential equation is (x+y)y’-x+y = 0 and the one parameter family of solutions
to verify is y^2+2xy-x^2 = C, alright, similar to the way we verified particular solutions,
we need to find the derivative of this general solution implicitly. For this differential
equation we need to find the first derivative. So let’s go ahead and take the derivative
of this solution implicitly, taking the derivative of each term with respect to x, and applying
the power rule along with the product rule, we obtain the following expression.
Having found the derivative implicitly we now need to solve for y prime, so let’s
go ahead and move all the terms containing y prime to one side and the rest of the terms
to the other side of the equation. Now that we have all the terms containing y prime on
one side we go ahead and factor it out as follows. Then it’s just a matter of solving
for y prime, so we divide both sides by the quantity 2y + 2x, finally we can simplify this
expression by factoring out a 2 in the numerator and denominator which simplifies to one.
In the end the first derivative of this implicit function is equal to the following expression.
Having found the first derivative we proceed with the substitution step, and verify that
both the left and right hand side of the differential equation are equal to each other. Substituting
the first derivative we obtain the following. Next let's go ahead and simplify the left hand
side of the equation by distributing the quantity x+y, then it’s just a matter of collecting
like terms, notice that all the terms cancel each other out, in the end we are left with the expression
0 = 0 this means that this one-parameter family of solutions is indeed a general solution
of this ODE.
Keep in mind that the interval of definition of this solution is dependent on the value
of the parameter c, in a later video we will be asked to find a particular solution of
this ODE, for now we only want to verify the family of solutions. Alright let’s try the final example.
The differential equation is y’ = [y(y^3-2x^3)/[x(2y^3-x^3)] and the one parameter family of solutions to verify
is x^3 + y^3 = 3cxy, alright let's go ahead and find the first derivative, we are going to
find the derivative by implicit differentiation, before we proceed let's first rearrange the
solution so we can verify it with the least effort possible, at first glance you
might be tempted to find the derivative implicitly in its current form. Technically there is
nothing stopping you in carrying out the differentiation this way, but notice that if you do happen
to take the derivative in its current form the parameter c will be tagging along throughout
the verification process. In the end this would add additional steps which can be avoided
by modifying the solution so that the parameter c is one side of the equation, we accomplish
this by dividing both sides of the solution by the expression x times y as follows.
This way when we carry out the differentiation step this parameter will be eliminated since
the derivative of a constant is just equal to zero. When taking derivatives implicitly
for the purpose of verifying a solution it’s always a good idea to try to eliminate the
parameter whenever possible.
Alright taking the derivative implicitly with respect to x requires an application of the
quotient rule. Applying the quotient rule results in the following expression. Next
let’s go ahead and simplify the expression in the numerator, notice that we can eliminate
the denominator by multiplying both sides by x squared times y squared. This leaves
us with the expression in the numerator to deal with, we first need to distribute the
factor x times y, and foil the quantity x^3 + y^3 with the quantity x times y prime + y,
carrying out he distributions we end up with the following terms. Next let’s distribute
the negative sign, and collect like terms, the main goal is to solve for the derivative
in this case y prime, so we need to move all the terms containing y prime to one side and
the rest of the terms to the other side of the equation, as follows. At this point it
is just a matter of factoring out the y prime term, and isolate it by dividing both sides
by the quantity (2xy^3-x^4), in the end we obtain the following expression for the first derivative.
Now let’s proceed with the substitution step, substituting the first derivative into
the differential equation results in the following expression, notice that we can factor out
a y term in the numerator and factor out the variable x in the denominator, lastly we can
rearrange the factor in the numerator so that it matches the right hand side of the equation.
Since both sides are equivalent we conclude that this one parameter family of solutions
is indeed a general solution to the differential equation. Alright in our next video
we go over how to verify a piece wise defined function constructed from a general solution.