Solving $frac{mathrm dy}{mathrm dx}+frac y{x^2}=frac1{x^2}$
$begingroup$
Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$
My Attempt
Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.
Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$
ordinary-differential-equations
edited Dec 26 '18 at 15:16
Saad
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
add a comment |
$begingroup$
Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$
My Attempt
Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.
Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$
ordinary-differential-equations
edited Dec 26 '18 at 15:16
Saad
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
6
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
5
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20
add a comment |
$begingroup$
Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$
My Attempt
Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.
Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$
ordinary-differential-equations
edited Dec 26 '18 at 15:16
Saad
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
$endgroup$
Solve the following equation:
$$
dfrac {dy}{dx} + dfrac {y}{x^2}=dfrac {1}{x^2}
$$
My Attempt
Given:
$$
dfrac {dy}{dx}+dfrac {y}{x^2}=dfrac {1}{x^2}
$$
Comparing above equation with the standard form of first order linear differential equation $dy/dx+P.y=Q$ where $P$ and $Q$ are the functions of $x$ or constants.
Now, using the integrating factor $e^{int P dx}$:
$$
e^{int P dx}=e^{int x^{-2} dx}=e^{-frac {1}{x}}
$$
Multiplying both sides of the given equation by integrating factor:
$$
e^{-frac {1}{x}}.dfrac {dy}{dx}+dfrac {y.e^{-frac {1}{x}}}{x^2}=dfrac {e^{-frac {1}{x}}}{x^2}
$$
ordinary-differential-equations
ordinary-differential-equations
edited Dec 26 '18 at 15:16
Saad
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
edited Dec 26 '18 at 15:16
Saad
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
edited Dec 26 '18 at 15:16
Saad
19.7k92352
edited Dec 26 '18 at 15:16
Saad
19.7k92352
edited Dec 26 '18 at 15:16
Saad
19.7k92352
19.7k92352
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
asked Dec 26 '18 at 13:14
blue_eyed_...blue_eyed_...
3,26721750
3,26721750
6
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
5
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20
add a comment |
6
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
5
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20
6
6
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
5
5
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
Great, now write it in the form
$$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$
We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.
Hence $$frac{partial F}{partial y} = e^{-1/x}$$
$$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$
Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.
Hence our equation is $$dF(x,y) = 0$$
so $F(x,y) = C$ for some $C in mathbb{R}$. We get
$$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
$begingroup$
This is a linear DE so considering instead
$$
x^2y'+y=1
$$
$$
y = y_h + y_p
$$
such that
$$
x^2y_h'+y_h = 0\
x^2y_p' + y_p = 1
$$
for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives
$$
-C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
$$
and for the particular $y_p = 1$ then
$$
y = C_0 e^{frac 1x}+1
$$
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
add a comment |
$begingroup$
Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
$endgroup$
Hint: Write your equation in the form
$$frac{dy}{1-y}=frac{dx}{x^2}$$ for $$yne 1$$ and $$xne 0$$
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
answered Dec 26 '18 at 13:29
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
76.2k42866
76.2k42866
add a comment |
add a comment |
$begingroup$
Great, now write it in the form
$$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$
We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.
Hence $$frac{partial F}{partial y} = e^{-1/x}$$
$$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$
Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.
Hence our equation is $$dF(x,y) = 0$$
so $F(x,y) = C$ for some $C in mathbb{R}$. We get
$$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
Great, now write it in the form
$$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$
We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.
Hence $$frac{partial F}{partial y} = e^{-1/x}$$
$$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$
Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.
Hence our equation is $$dF(x,y) = 0$$
so $F(x,y) = C$ for some $C in mathbb{R}$. We get
$$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
Great, now write it in the form
$$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$
We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.
Hence $$frac{partial F}{partial y} = e^{-1/x}$$
$$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$
Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.
Hence our equation is $$dF(x,y) = 0$$
so $F(x,y) = C$ for some $C in mathbb{R}$. We get
$$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
$endgroup$
Great, now write it in the form
$$frac{e^{-1/x}}{x^2}(y-1),dx + e^{-1/x} ,dy = 0$$
We are trying to find a function $F(x,y)$ such that the LHS above is the total differential of $F$.
Hence $$frac{partial F}{partial y} = e^{-1/x}$$
$$frac{partial F}{partial y} = frac{e^{-1/x}}{x^2}(y-1)$$
Solving this we get e.g. $F(x,y) = e^{-1/x}(y-1)$.
Hence our equation is $$dF(x,y) = 0$$
so $F(x,y) = C$ for some $C in mathbb{R}$. We get
$$e^{-1/x}(y-1) = C implies y = 1+Ce^{1/x}$$
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
answered Dec 26 '18 at 13:33
mechanodroidmechanodroid
27.8k62447
27.8k62447
add a comment |
add a comment |
$begingroup$
I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
$begingroup$
I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
$endgroup$
add a comment |
$begingroup$
I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
$endgroup$
I'll review the general solution from an integration factor. Let $R:=expint Pdx$ so $R'=RP,,(Ry)'=R(y'+Py)=RQ,,y=R^{-1}int RQdx$. In the special case $P=Q$, as in this problem, we in fact have $RQ=R'$ so $y=1+CR^{-1}$. As others have noted, $P=Q$ also makes the equation separable, viz. $dy/(1-y)=Pdx$ so $-ln |1-y|=ln R+C$, i.e. $|1-y|propto R^{-1}$.
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
answered Dec 26 '18 at 13:44
J.G.J.G.
27.8k22843
27.8k22843
add a comment |
add a comment |
$begingroup$
This is a linear DE so considering instead
$$
x^2y'+y=1
$$
$$
y = y_h + y_p
$$
such that
$$
x^2y_h'+y_h = 0\
x^2y_p' + y_p = 1
$$
for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives
$$
-C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
$$
and for the particular $y_p = 1$ then
$$
y = C_0 e^{frac 1x}+1
$$
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
$endgroup$
add a comment |
$begingroup$
This is a linear DE so considering instead
$$
x^2y'+y=1
$$
$$
y = y_h + y_p
$$
such that
$$
x^2y_h'+y_h = 0\
x^2y_p' + y_p = 1
$$
for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives
$$
-C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
$$
and for the particular $y_p = 1$ then
$$
y = C_0 e^{frac 1x}+1
$$
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
$endgroup$
add a comment |
$begingroup$
This is a linear DE so considering instead
$$
x^2y'+y=1
$$
$$
y = y_h + y_p
$$
such that
$$
x^2y_h'+y_h = 0\
x^2y_p' + y_p = 1
$$
for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives
$$
-C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
$$
and for the particular $y_p = 1$ then
$$
y = C_0 e^{frac 1x}+1
$$
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
$endgroup$
This is a linear DE so considering instead
$$
x^2y'+y=1
$$
$$
y = y_h + y_p
$$
such that
$$
x^2y_h'+y_h = 0\
x^2y_p' + y_p = 1
$$
for the homogeneous solution $y_h = C_0 e^{frac 1x}$ which after substitution gives
$$
-C_0frac{x^2}{x^2}e^{frac 1x}+C_0 e^{frac 1x}=0
$$
and for the particular $y_p = 1$ then
$$
y = C_0 e^{frac 1x}+1
$$
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
edited Dec 26 '18 at 14:53
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
answered Dec 26 '18 at 13:50
CesareoCesareo
8,9913516
8,9913516
add a comment |
add a comment |
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6
$begingroup$
OK. And the next step is to recognize the left-hand side as the derivative of ...something, something that's typically a product. For then you'll have $f'(x) = $ right-hand-side, and you can find $f$ by integration. So what is the left-hand-side the derivative of?
$endgroup$
– John Hughes
Dec 26 '18 at 13:18
5
$begingroup$
@blue_eyed_: This is a Separable Equation.
$endgroup$
– Moo
Dec 26 '18 at 13:20