General solution of $xy '= y+sqrt{x^2+y^2}$
$begingroup$
Find the general solution of the folowing differential equation:
$$xy '= y+sqrt{x^2+y^2}$$
I try to solve it
this my result: $sqrt{v^2 +1}+v=x+c$ where $c$ is a constant.
I don't know if its correct or not !
calculus ordinary-differential-equations
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
$endgroup$
add a comment |
$begingroup$
Find the general solution of the folowing differential equation:
$$xy '= y+sqrt{x^2+y^2}$$
I try to solve it
this my result: $sqrt{v^2 +1}+v=x+c$ where $c$ is a constant.
I don't know if its correct or not !
calculus ordinary-differential-equations
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
$endgroup$
add a comment |
$begingroup$
Find the general solution of the folowing differential equation:
$$xy '= y+sqrt{x^2+y^2}$$
I try to solve it
this my result: $sqrt{v^2 +1}+v=x+c$ where $c$ is a constant.
I don't know if its correct or not !
calculus ordinary-differential-equations
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
$endgroup$
Find the general solution of the folowing differential equation:
$$xy '= y+sqrt{x^2+y^2}$$
I try to solve it
this my result: $sqrt{v^2 +1}+v=x+c$ where $c$ is a constant.
I don't know if its correct or not !
calculus ordinary-differential-equations
calculus ordinary-differential-equations
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
edited Dec 9 '18 at 20:44
quid♦
37.1k95093
37.1k95093
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
asked Dec 9 '18 at 18:27
hmeteirhmeteir
193
193
add a comment |
add a comment |
1 Answer
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Hint: Let $$y(x)=xcdot v(x)$$ then you will get
$$frac{dv(x)}{dx}=frac{sqrt{v(x)^2+1}}{x}$$
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
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add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Hint: Let $$y(x)=xcdot v(x)$$ then you will get
$$frac{dv(x)}{dx}=frac{sqrt{v(x)^2+1}}{x}$$
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
add a comment |
$begingroup$
Hint: Let $$y(x)=xcdot v(x)$$ then you will get
$$frac{dv(x)}{dx}=frac{sqrt{v(x)^2+1}}{x}$$
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
add a comment |
$begingroup$
Hint: Let $$y(x)=xcdot v(x)$$ then you will get
$$frac{dv(x)}{dx}=frac{sqrt{v(x)^2+1}}{x}$$
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
Hint: Let $$y(x)=xcdot v(x)$$ then you will get
$$frac{dv(x)}{dx}=frac{sqrt{v(x)^2+1}}{x}$$
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Dec 9 '18 at 19:10
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
75.4k42865
add a comment |
add a comment |
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