Define the term homogeneous first order ordinary differential equation. Hence solve x^2 y' + xy = x2 + y

Rui Baldwin 2021-02-09 Answered
Define the term homogeneous first order ordinary differential equation.
Hence solve x2y+xy=x2+y
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Maciej Morrow
Answered 2021-02-10 Author has 98 answers

Now solve the given differential equation
Thus, the solution of differential equation is
Solving further to get
Ei(x) is the exponential integral

Not exactly what you’re looking for?
Ask My Question
Jeffrey Jordon
Answered 2021-10-23 Author has 2070 answers

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2022-01-21
Are the following are linear equation?
1. sin(x)dydx3y=0
2. d2ydx2+sin(x)dydx=cos(x)
asked 2022-02-15
Suppose that a dose of x0 (initial) gram of a drug is injected into the bloodstream. Assume that the drug leaves the blood and enters the urine at a rate proportional to the amount of drug present in the blood. In addition, assume that half of the drug dose has entered the urine after 0.75 hour. Find the time at which the amount of drug in the blood stream is 5% of the original drug dose x0 (initial).
I have the first order differential equation. It's finding t that is the problem.
asked 2022-01-22
Write an equivalent first-order differential equation and initial condition for y.
What is the equivalent first-order differential equation?
What is the initial condition?
asked 2022-01-22
Solve the equation separable, linear, bernoulli, or homogenous
asked 2022-05-27
If I have a differential equation y ( t ) = A y ( t ) where A is a constant square matrix that is not diagonalizable(although it is surely possible to calculate the eigenvalues) and no initial condition is given. And now I am interested in the fundamental matrix. Is there a general method to determine this matrix? I do not want to use the exponential function and the Jordan normal form, as this is quite exhausting. Maybe there is also an ansatz possible as it is for the special case, where this differential equation is equivalent to an n-th order ode. I saw a method where they calculated the eigenvalues of the matrix and depending on the multiplicity n of this eigenvalue they used an exponential term(with the eigenvalue) and in each component an n-th order polynomial as a possible ansatz. Though they only did this, when they were interested in a initial value problem, so with an initial condition and not for a general solution.
I was asked to deliver an example: so y ( t ) = ( 3 4 1 1 ) y ( t ) If somebody can construct a fundamental matrix for this system, than this should be sufficient
asked 2022-02-16
I've tried many times to reach the solution of a first order differential equation (of the last equation) but unfortunately I couldn't. Could you please help me to know how did he get this solution.
The equation is:
The author assumed that
X=0 at t=0
Then, he said in his first publication the solution is given as:
I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.
In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:
X=kf(kf+kb){1exp[(kf+kb)t]}+X0 exp[(kf+kb)t] and he said X0 was taken to be zero in the first publication.
Can any of you help how did the author get this solution please?
asked 2022-04-10
For every differentiable function f : R R , there is a function g : R × R R such that g ( f ( x ) , f ( x ) ) = 0 for every x and for every differentiable function h : R R holds that
being true that for every x R , g ( h ( x ) , h ( x ) ) = 0 and h ( 0 ) = f ( 0 ) implies that h ( x ) = f ( x ) for every x R .
i.e every differentiable function f is a solution to some first order differential equation that has translation symmetry.