What is a solution to the differential equation dy/dx=x+y?

Joyce Sharp 2022-09-18 Answered
What is a solution to the differential equation d y d x = x + 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

Answers (1)

Nathalie Rivers
Answered 2022-09-19 Author has 7 answers
Let u=x+y
d u d x = d d x ( x + y ) = 1 + d y d x
d y d x = d u d x - 1
Thus, making the substitutions into our original equation,
d u d x - 1 = u
d u u + 1 = d x
d u u + 1 = d x
ln ( u + 1 ) = x + C 0
e ln ( u + 1 ) = e x + C 0
u + 1 = C 1 e x (where C 1 = e C 0 )
Substituting x+y=u back in,
x + y + 1 = C 1 e x
y = C 1 e x - x - 1

We have step-by-step solutions for your answer!

Expert Community at Your Service

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

You might be interested in

asked 2022-01-20
Who know?
Solve the following first order differential equations. Use any method:
2x2dx3xy2dy=0; y(1)=0
asked 2020-12-29
Solve differential equationxy+[(2x+1)/(x+1)]y=x1
asked 2022-06-12
Let the first order differential equation be
d y / d x + P ( x ) y = Q ( x )
To solve this we multiply it by function say v(x) then it becomes
v ( x ) d y / d x + P ( x ) v ( x ) y = Q ( x ) v ( x )
d ( v ( x ) y ) / d x = Q ( x ) v ( x )
where d ( v ( x ) ) / d x = P ( x ) v ( x )
I don't write full solution here but now here to obtain integrating factor v(x) from above equation we get l n | v ( x ) | = P ( x ) d x.
Now my question is why we neglect absolute value of v(x) and obtain integrating factor as e P ( x ) d x = v ( x )
asked 2020-11-07
Solve differential equation xydyy2dx=(x+y)2e(y/x)
asked 2022-05-28
I would need some help in solving the following differential equation:
u ( p ) + r + N λ p N λ p ( 1 p ) u ( p ) = r + N λ p N λ ( 1 p ) g .
I already know that the solution is of the form
g p + C ( 1 p ) ( 1 p p ) r / λ N ,,
where C is a constant.
Now this is a standard linear first-order differential equation. Letting
f ( p ) = r + N λ p N λ p ( 1 p ) ,     q ( p ) = r + N λ p N λ ( 1 p ) g ,     μ ( p ) = e f ( p ) d p ,
the solution should have the form
u ( p ) = C μ ( p ) + 1 μ ( p ) μ ( p ) q ( p ) d p ,
where C is a constant.
The calculation of μ ( p ) was not much of a problem. Indeed, we can rewrite f(p) as
f ( p ) = r N λ 1 p + ( 1 + r N λ ) 1 1 p
Using 1 p d p = ln ( p ) and 1 1 p d p = ln ( 1 p ), I found that
μ ( p ) = p r N λ ( 1 p ) 1 r N λ .
Where I have difficulties is calculating the term ∫μ(p)q(p)dp, where
μ ( p ) q ( p ) = r g N λ p r N λ ( 1 p ) 2 r N λ + p r N λ + 1 ( 1 p ) 2 r N λ .
How do I integrate this last term? Or did I miss something obvious, because in the paper I'm looking at they don't provide any details about how they get to the solution of the differential equation
Edit: small typo corrected in the last displayed equation.
Edit: there is a big typo in the differential equation: the rhs should have r + N λ rather than r + N λ p in the numerator. This probably makes it much simpler to solve. My bad.
asked 2022-05-22
t y + t y = 1 y
I am having trouble going about this question. I have tried different processes to answer the differential equation, including separable, exact, homogeneous and linear equations. I have ran into a wall through all these processes. Maybe I'm not following the correct procedure... Help!
I know the first step would be to divide both sides by t, which would give:
y + y = ( 1 y ) / t
asked 2021-02-24
Solve the differential equation xdy/dx=y+xe(y/x), y=vx

New questions