Thread: differential equations

show that y=e^-t is a solution of d^2y/dt^2-dy/dt - 2y=0.

find the values of m for which y=e^mt is a solution of d^2y/dt^2+dy/dt -6y=0.

find the solution of the first order initial- value problem dy/dx=2x. y(1)=2.

Solve the initia-value problem d^2y/dx^2+ dy/dx = e^x y(0) = 1, y (0)=0.

Originally Posted by lay137

show that y=e^-t is a solution of d^2y/dt^2-dy/dt - 2y=0.

Simplify substitute or you can write the charachteristic equation:
k^2-k-2=0
Thus, k=2,-1
Thus, the set
C_1*e^{2t}+C_2*e^{-t} is the basis for the solutions.

find the values of m for which y=e^mt is a solution of d^2y/dt^2+dy/dt -6y=0.

Again, charachterstic equation,
k^2+k-6=0
Thus,
k=-3,2
Thus,
e^{-3t} and e^{2t} are solutions,
Thus possible values of "m" are,
-3,2

Originally Posted by lay137

find the solution of the first order initial- value problem dy/dx=2x. y(1)=2.

You have,
y'=2x
Simpligy integrate, all solutions form the set,
y=x^2+C
When x=2 then y=1 (initial value),
1=4+C thus, C=-3
Thus,
y=x^2-3
Is the unique solution to this initial problem.

Solve the initia-value problem d^2y/dx^2+ dy/dx = e^x y'(0) = 1, y (0)=0.

The homogenous equation is,
y''+y'=e^x
Reduction of order let u=y' then u'=y''
Thus,
u'+u=e^x
This is a first order linear differencial equation.
Thus,
u=1/m(x) * INTEGRAL m(x)e^x dx
Where m(x) is integrating factor which is,
m(x)=exp (INTEGRAL 1 dx)=e^x
Thus, we have,
u=e^{-x}*INTEGRAL e^{2x}dx
Thus,
u=e^{-x}*[(1/2)e^{2x}+C]
Thus,
u=(1/2)e^x+C*e^{-x}
Then y is the integral of that,
y=(1/2)e^x-C*e^{-x}+K

We have from all of this that,
y'=(1/2)e^x+C*e^{-x}
y=(1/2)e^x-C*e^{-x}+K
Now we substitute initial conditions,
1=(1/2)+C thus, C=1/2
0=(1/2)-C+K thus, K=0
Therefore the unique solution is,
y=(1/2)e^x+(1/2)*e^{-x}