Solve the differential equation $frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$
$begingroup$
Solve the differential equation $$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$frac{dy}{dx}=frac{1}{y^3}frac{left(1+2y^4right)}{1+frac{4x}{y^4}}$$
Multiplying both sides with $frac{1}{y^5}$ we get:
$$frac{1}{y^5}frac{dy}{dx}=frac{1}{y^8}frac{y^4(2+frac{1}{y^4})}{1+frac{4x}{y^4}}=frac{1}{y^4}frac{(2+frac{1}{y^4})}{1+frac{4x}{y^4}}$$
Now letting $$frac{1}{y^4}=t$$ we get
$$frac{-1}{4}frac{dt}{dx}=frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
algebra-precalculus ordinary-differential-equations homogeneous-equation
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
$endgroup$
add a comment |
$begingroup$
Solve the differential equation $$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$frac{dy}{dx}=frac{1}{y^3}frac{left(1+2y^4right)}{1+frac{4x}{y^4}}$$
Multiplying both sides with $frac{1}{y^5}$ we get:
$$frac{1}{y^5}frac{dy}{dx}=frac{1}{y^8}frac{y^4(2+frac{1}{y^4})}{1+frac{4x}{y^4}}=frac{1}{y^4}frac{(2+frac{1}{y^4})}{1+frac{4x}{y^4}}$$
Now letting $$frac{1}{y^4}=t$$ we get
$$frac{-1}{4}frac{dt}{dx}=frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
algebra-precalculus ordinary-differential-equations homogeneous-equation
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
$endgroup$
add a comment |
$begingroup$
Solve the differential equation $$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$frac{dy}{dx}=frac{1}{y^3}frac{left(1+2y^4right)}{1+frac{4x}{y^4}}$$
Multiplying both sides with $frac{1}{y^5}$ we get:
$$frac{1}{y^5}frac{dy}{dx}=frac{1}{y^8}frac{y^4(2+frac{1}{y^4})}{1+frac{4x}{y^4}}=frac{1}{y^4}frac{(2+frac{1}{y^4})}{1+frac{4x}{y^4}}$$
Now letting $$frac{1}{y^4}=t$$ we get
$$frac{-1}{4}frac{dt}{dx}=frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
algebra-precalculus ordinary-differential-equations homogeneous-equation
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
$endgroup$
Solve the differential equation $$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
My try:
we can write the equation as:
$$frac{dy}{dx}=frac{1}{y^3}frac{left(1+2y^4right)}{1+frac{4x}{y^4}}$$
Multiplying both sides with $frac{1}{y^5}$ we get:
$$frac{1}{y^5}frac{dy}{dx}=frac{1}{y^8}frac{y^4(2+frac{1}{y^4})}{1+frac{4x}{y^4}}=frac{1}{y^4}frac{(2+frac{1}{y^4})}{1+frac{4x}{y^4}}$$
Now letting $$frac{1}{y^4}=t$$ we get
$$frac{-1}{4}frac{dt}{dx}=frac{t^2+2t}{4tx+1}$$
Any way further to convert in to variable separable?
algebra-precalculus ordinary-differential-equations homogeneous-equation
algebra-precalculus ordinary-differential-equations homogeneous-equation
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
asked Dec 21 '18 at 5:30
Umesh shankarUmesh shankar
2,81631220
2,81631220
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
This is an "almost" exact equation. Write it in $A(x,y)text dx + B(x,y)text dy=0$ form
$$(-y-2y^5)text dx + (4x+y^4)text dy = 0$$
Then, look for a function $mu(y)$ such that
$$ (-y-2y^5)mu(y)text dx + (4x+y^4)mu(y)text dy = 0 $$
Let $mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)text dx + (4xy^{-5}+y^{-1})text dy = 0 \
text dleft[ -xy^{-4} - 2x + log y right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + log y = C $$
Solving for $x$ gives us
$$ x = frac{y^4}{2y^4+1}log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)exp(W(z)) = W(zexp(z)) = z$,
$$y = left( frac{4x}{W(kxexp(-8x))} right)^frac{1}{4}$$
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=frac{y^4}{2y^4+1}ln(c:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=left(frac{4x}{text{W}(C:xe^{-8x})}right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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$begingroup$
This is an "almost" exact equation. Write it in $A(x,y)text dx + B(x,y)text dy=0$ form
$$(-y-2y^5)text dx + (4x+y^4)text dy = 0$$
Then, look for a function $mu(y)$ such that
$$ (-y-2y^5)mu(y)text dx + (4x+y^4)mu(y)text dy = 0 $$
Let $mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)text dx + (4xy^{-5}+y^{-1})text dy = 0 \
text dleft[ -xy^{-4} - 2x + log y right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + log y = C $$
Solving for $x$ gives us
$$ x = frac{y^4}{2y^4+1}log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)exp(W(z)) = W(zexp(z)) = z$,
$$y = left( frac{4x}{W(kxexp(-8x))} right)^frac{1}{4}$$
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
add a comment |
$begingroup$
This is an "almost" exact equation. Write it in $A(x,y)text dx + B(x,y)text dy=0$ form
$$(-y-2y^5)text dx + (4x+y^4)text dy = 0$$
Then, look for a function $mu(y)$ such that
$$ (-y-2y^5)mu(y)text dx + (4x+y^4)mu(y)text dy = 0 $$
Let $mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)text dx + (4xy^{-5}+y^{-1})text dy = 0 \
text dleft[ -xy^{-4} - 2x + log y right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + log y = C $$
Solving for $x$ gives us
$$ x = frac{y^4}{2y^4+1}log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)exp(W(z)) = W(zexp(z)) = z$,
$$y = left( frac{4x}{W(kxexp(-8x))} right)^frac{1}{4}$$
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
add a comment |
$begingroup$
This is an "almost" exact equation. Write it in $A(x,y)text dx + B(x,y)text dy=0$ form
$$(-y-2y^5)text dx + (4x+y^4)text dy = 0$$
Then, look for a function $mu(y)$ such that
$$ (-y-2y^5)mu(y)text dx + (4x+y^4)mu(y)text dy = 0 $$
Let $mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)text dx + (4xy^{-5}+y^{-1})text dy = 0 \
text dleft[ -xy^{-4} - 2x + log y right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + log y = C $$
Solving for $x$ gives us
$$ x = frac{y^4}{2y^4+1}log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)exp(W(z)) = W(zexp(z)) = z$,
$$y = left( frac{4x}{W(kxexp(-8x))} right)^frac{1}{4}$$
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
$endgroup$
This is an "almost" exact equation. Write it in $A(x,y)text dx + B(x,y)text dy=0$ form
$$(-y-2y^5)text dx + (4x+y^4)text dy = 0$$
Then, look for a function $mu(y)$ such that
$$ (-y-2y^5)mu(y)text dx + (4x+y^4)mu(y)text dy = 0 $$
Let $mu(y) = y^{-5}$. Then, the equation becomes
$$ (-y^{-4}-2)text dx + (4xy^{-5}+y^{-1})text dy = 0 \
text dleft[ -xy^{-4} - 2x + log y right] = 0$$
Integrating, we get
$$ -xy^{-4} - 2x + log y = C $$
Solving for $x$ gives us
$$ x = frac{y^4}{2y^4+1}log(cy) $$
Solving for $y$ in terms of the function $W(z)$ such that $W(z)exp(W(z)) = W(zexp(z)) = z$,
$$y = left( frac{4x}{W(kxexp(-8x))} right)^frac{1}{4}$$
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
edited Dec 21 '18 at 8:09
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
answered Dec 21 '18 at 7:43
AlexanderJ93AlexanderJ93
6,173823
6,173823
add a comment |
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=frac{y^4}{2y^4+1}ln(c:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=left(frac{4x}{text{W}(C:xe^{-8x})}right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=frac{y^4}{2y^4+1}ln(c:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=left(frac{4x}{text{W}(C:xe^{-8x})}right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
$endgroup$
add a comment |
$begingroup$
$$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=frac{y^4}{2y^4+1}ln(c:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=left(frac{4x}{text{W}(C:xe^{-8x})}right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
$endgroup$
$$frac{dy}{dx}=frac{y+2y^5}{4x+y^4}$$
$$(y+2y^5)frac{dx}{dy}-4x=y^4$$
Considering the function $x(y)$ this is a first order linear ODE. Solving such kind of ODE is classic. The result is :
$$x(y)=frac{y^4}{2y^4+1}ln(c:y)$$
$y(x)$ is the inverse function. It cannot be expressed with a finite number of elementary functions. A closed form requires a special function W(X) the Lambert W function.
$$y(x)=left(frac{4x}{text{W}(C:xe^{-8x})}right)^{1/4}$$
$C$ is an arbitrary constant, to be determined according to some boundary condition.
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
edited Dec 21 '18 at 7:52
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
answered Dec 21 '18 at 7:46
JJacquelinJJacquelin
44.2k21854
44.2k21854
add a comment |
add a comment |
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