Dividing both sides of a differential equation by a variable.Can anyone explain the intuitive meaning of 'integrating on both sides of the equation' when solving differential equations?Differential Equation - how to solve for the equation of a curve passing through points?Finding the equation of the curve. (Differential Equation)Differential Equation: Initial Value Problemdifferential equation with linear coefficientsIntegrating on both sides - differential equation - intuitionDifferential Equations-Variable Seperable MethodHow do I apply integrating factor to solve this differential equation?Confusion about Global solution and Local solution of a first order differential equationHow can I know how many solutions of a differential equation passing a point?
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Dividing both sides of a differential equation by a variable.
Can anyone explain the intuitive meaning of 'integrating on both sides of the equation' when solving differential equations?Differential Equation - how to solve for the equation of a curve passing through points?Finding the equation of the curve. (Differential Equation)Differential Equation: Initial Value Problemdifferential equation with linear coefficientsIntegrating on both sides - differential equation - intuitionDifferential Equations-Variable Seperable MethodHow do I apply integrating factor to solve this differential equation?Confusion about Global solution and Local solution of a first order differential equationHow can I know how many solutions of a differential equation passing a point?
$begingroup$
The question specifically states-
Find the equation of a curve passing through the point $(0,1)$ given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of $x$-coordinate and the product of the $x$-coordinate and $y$-coordinate of the point.
My solution:
$$
fracdydx= x + xy
$$
or, $frac1x (fracdydx)= 1 + y$ [ dividing both sides by x]
or, $frac11+ydy$= $x$ $dx$
now integrating both sides
$log(1+y)= x^2/2 + C$
But my general eqn of the curve seems to be wrong. I think I made the mistake while dividing by x.
I know about dividing both sides of an algebraic equation by a variable and the precautions needed there. But I'm new to differential equations and not sure how dividing by variables work here. I have, however, solved a previous problem by dividing by a variable, namely-
$fracdydx$=$frac2x-yx+y$
[ I divided numerator and denominator of RHS by x and and put y/x=v]
But the first problem, I couldn't solve by diving variables. So... why?
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
ordinary-differential-equations
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
$endgroup$
add a comment |
$begingroup$
The question specifically states-
Find the equation of a curve passing through the point $(0,1)$ given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of $x$-coordinate and the product of the $x$-coordinate and $y$-coordinate of the point.
My solution:
$$
fracdydx= x + xy
$$
or, $frac1x (fracdydx)= 1 + y$ [ dividing both sides by x]
or, $frac11+ydy$= $x$ $dx$
now integrating both sides
$log(1+y)= x^2/2 + C$
But my general eqn of the curve seems to be wrong. I think I made the mistake while dividing by x.
I know about dividing both sides of an algebraic equation by a variable and the precautions needed there. But I'm new to differential equations and not sure how dividing by variables work here. I have, however, solved a previous problem by dividing by a variable, namely-
$fracdydx$=$frac2x-yx+y$
[ I divided numerator and denominator of RHS by x and and put y/x=v]
But the first problem, I couldn't solve by diving variables. So... why?
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
ordinary-differential-equations
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
$endgroup$
$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
sorry, changed now
$endgroup$
– Anurag Saha
Mar 12 at 8:29
1
$begingroup$
What is the solution in the textbook? Maybe you have an equivalent expresssion.
$endgroup$
– Paras Khosla
Mar 12 at 8:38
add a comment |
$begingroup$
The question specifically states-
Find the equation of a curve passing through the point $(0,1)$ given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of $x$-coordinate and the product of the $x$-coordinate and $y$-coordinate of the point.
My solution:
$$
fracdydx= x + xy
$$
or, $frac1x (fracdydx)= 1 + y$ [ dividing both sides by x]
or, $frac11+ydy$= $x$ $dx$
now integrating both sides
$log(1+y)= x^2/2 + C$
But my general eqn of the curve seems to be wrong. I think I made the mistake while dividing by x.
I know about dividing both sides of an algebraic equation by a variable and the precautions needed there. But I'm new to differential equations and not sure how dividing by variables work here. I have, however, solved a previous problem by dividing by a variable, namely-
$fracdydx$=$frac2x-yx+y$
[ I divided numerator and denominator of RHS by x and and put y/x=v]
But the first problem, I couldn't solve by diving variables. So... why?
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
ordinary-differential-equations
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
$endgroup$
The question specifically states-
Find the equation of a curve passing through the point $(0,1)$ given that the slope of the tangent to the curve at any point $(x,y)$ is equal to the sum of $x$-coordinate and the product of the $x$-coordinate and $y$-coordinate of the point.
My solution:
$$
fracdydx= x + xy
$$
or, $frac1x (fracdydx)= 1 + y$ [ dividing both sides by x]
or, $frac11+ydy$= $x$ $dx$
now integrating both sides
$log(1+y)= x^2/2 + C$
But my general eqn of the curve seems to be wrong. I think I made the mistake while dividing by x.
I know about dividing both sides of an algebraic equation by a variable and the precautions needed there. But I'm new to differential equations and not sure how dividing by variables work here. I have, however, solved a previous problem by dividing by a variable, namely-
$fracdydx$=$frac2x-yx+y$
[ I divided numerator and denominator of RHS by x and and put y/x=v]
But the first problem, I couldn't solve by diving variables. So... why?
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
ordinary-differential-equations
ordinary-differential-equations
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
edited Mar 12 at 9:04
Anurag Saha
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
asked Mar 12 at 8:23
Anurag SahaAnurag Saha
386
386
$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
sorry, changed now
$endgroup$
– Anurag Saha
Mar 12 at 8:29
1
$begingroup$
What is the solution in the textbook? Maybe you have an equivalent expresssion.
$endgroup$
– Paras Khosla
Mar 12 at 8:38
add a comment |
$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
sorry, changed now
$endgroup$
– Anurag Saha
Mar 12 at 8:29
1
$begingroup$
What is the solution in the textbook? Maybe you have an equivalent expresssion.
$endgroup$
– Paras Khosla
Mar 12 at 8:38
$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
sorry, changed now
$endgroup$
– Anurag Saha
Mar 12 at 8:29
$begingroup$
sorry, changed now
$endgroup$
– Anurag Saha
Mar 12 at 8:29
1
1
$begingroup$
What is the solution in the textbook? Maybe you have an equivalent expresssion.
$endgroup$
– Paras Khosla
Mar 12 at 8:38
$begingroup$
What is the solution in the textbook? Maybe you have an equivalent expresssion.
$endgroup$
– Paras Khosla
Mar 12 at 8:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
What makes you think your solution is incorrect? As far I can tell, you did everything correctly. One way to confirm this is to note that you can change
$$log(1+y)= x^2/2 + C tag1labeleq1$$
by taking both sides to the power of $e$, to get
$$y + 1 = C_1 e^fracx^22 tag2labeleq2$$
where $C_1 = e^C$. Now, using the initial condition of $(0, 1)$, we get that
$$1 + 1 = C_1 e^0 Rightarrow C_1 = 2 tag3labeleq3$$
Thus, your solution would be
$$y = 2e^fracx^22 - 1 tag4labeleq4$$
If you wish to confirm the original differential equation works, if you differentiate each side wrt $x$, you get
$$fracdydx = 2left(frac12right)left(2xright)e^fracx^22 = 2xe^fracx^22 tag5labeleq5$$
Using that $y + 1 = 2e^fracx^22$ gives
$$fracdydx = xleft(y + 1right) = x + xy tag6labeleq6$$
This is your original equation.
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
add a comment |
$begingroup$
From $log(1+y)=fracx^22+C$, you get that $y=e^x^2/2+C-1$. Let us test this. You have then$$y'(x)=xe^x^2/2+C=xleft(e^x^2/2+C-1right)+x=xy(x)+x.$$So, yes, what you got is a solution.
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
What makes you think your solution is incorrect? As far I can tell, you did everything correctly. One way to confirm this is to note that you can change
$$log(1+y)= x^2/2 + C tag1labeleq1$$
by taking both sides to the power of $e$, to get
$$y + 1 = C_1 e^fracx^22 tag2labeleq2$$
where $C_1 = e^C$. Now, using the initial condition of $(0, 1)$, we get that
$$1 + 1 = C_1 e^0 Rightarrow C_1 = 2 tag3labeleq3$$
Thus, your solution would be
$$y = 2e^fracx^22 - 1 tag4labeleq4$$
If you wish to confirm the original differential equation works, if you differentiate each side wrt $x$, you get
$$fracdydx = 2left(frac12right)left(2xright)e^fracx^22 = 2xe^fracx^22 tag5labeleq5$$
Using that $y + 1 = 2e^fracx^22$ gives
$$fracdydx = xleft(y + 1right) = x + xy tag6labeleq6$$
This is your original equation.
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
add a comment |
$begingroup$
What makes you think your solution is incorrect? As far I can tell, you did everything correctly. One way to confirm this is to note that you can change
$$log(1+y)= x^2/2 + C tag1labeleq1$$
by taking both sides to the power of $e$, to get
$$y + 1 = C_1 e^fracx^22 tag2labeleq2$$
where $C_1 = e^C$. Now, using the initial condition of $(0, 1)$, we get that
$$1 + 1 = C_1 e^0 Rightarrow C_1 = 2 tag3labeleq3$$
Thus, your solution would be
$$y = 2e^fracx^22 - 1 tag4labeleq4$$
If you wish to confirm the original differential equation works, if you differentiate each side wrt $x$, you get
$$fracdydx = 2left(frac12right)left(2xright)e^fracx^22 = 2xe^fracx^22 tag5labeleq5$$
Using that $y + 1 = 2e^fracx^22$ gives
$$fracdydx = xleft(y + 1right) = x + xy tag6labeleq6$$
This is your original equation.
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
add a comment |
$begingroup$
What makes you think your solution is incorrect? As far I can tell, you did everything correctly. One way to confirm this is to note that you can change
$$log(1+y)= x^2/2 + C tag1labeleq1$$
by taking both sides to the power of $e$, to get
$$y + 1 = C_1 e^fracx^22 tag2labeleq2$$
where $C_1 = e^C$. Now, using the initial condition of $(0, 1)$, we get that
$$1 + 1 = C_1 e^0 Rightarrow C_1 = 2 tag3labeleq3$$
Thus, your solution would be
$$y = 2e^fracx^22 - 1 tag4labeleq4$$
If you wish to confirm the original differential equation works, if you differentiate each side wrt $x$, you get
$$fracdydx = 2left(frac12right)left(2xright)e^fracx^22 = 2xe^fracx^22 tag5labeleq5$$
Using that $y + 1 = 2e^fracx^22$ gives
$$fracdydx = xleft(y + 1right) = x + xy tag6labeleq6$$
This is your original equation.
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
$endgroup$
What makes you think your solution is incorrect? As far I can tell, you did everything correctly. One way to confirm this is to note that you can change
$$log(1+y)= x^2/2 + C tag1labeleq1$$
by taking both sides to the power of $e$, to get
$$y + 1 = C_1 e^fracx^22 tag2labeleq2$$
where $C_1 = e^C$. Now, using the initial condition of $(0, 1)$, we get that
$$1 + 1 = C_1 e^0 Rightarrow C_1 = 2 tag3labeleq3$$
Thus, your solution would be
$$y = 2e^fracx^22 - 1 tag4labeleq4$$
If you wish to confirm the original differential equation works, if you differentiate each side wrt $x$, you get
$$fracdydx = 2left(frac12right)left(2xright)e^fracx^22 = 2xe^fracx^22 tag5labeleq5$$
Using that $y + 1 = 2e^fracx^22$ gives
$$fracdydx = xleft(y + 1right) = x + xy tag6labeleq6$$
This is your original equation.
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
answered Mar 12 at 8:46
John OmielanJohn Omielan
3,9251215
3,9251215
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
add a comment |
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
1
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:05
add a comment |
$begingroup$
From $log(1+y)=fracx^22+C$, you get that $y=e^x^2/2+C-1$. Let us test this. You have then$$y'(x)=xe^x^2/2+C=xleft(e^x^2/2+C-1right)+x=xy(x)+x.$$So, yes, what you got is a solution.
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
add a comment |
$begingroup$
From $log(1+y)=fracx^22+C$, you get that $y=e^x^2/2+C-1$. Let us test this. You have then$$y'(x)=xe^x^2/2+C=xleft(e^x^2/2+C-1right)+x=xy(x)+x.$$So, yes, what you got is a solution.
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
add a comment |
$begingroup$
From $log(1+y)=fracx^22+C$, you get that $y=e^x^2/2+C-1$. Let us test this. You have then$$y'(x)=xe^x^2/2+C=xleft(e^x^2/2+C-1right)+x=xy(x)+x.$$So, yes, what you got is a solution.
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
$endgroup$
From $log(1+y)=fracx^22+C$, you get that $y=e^x^2/2+C-1$. Let us test this. You have then$$y'(x)=xe^x^2/2+C=xleft(e^x^2/2+C-1right)+x=xy(x)+x.$$So, yes, what you got is a solution.
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
answered Mar 12 at 8:41
José Carlos SantosJosé Carlos Santos
168k22132236
168k22132236
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
add a comment |
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
1
1
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
Update: My textbook used the method of Integrating factors and got a different solution making me think my solution was wrong. Turns out both versions are equivalent to each other and my solution was correct too.
$endgroup$
– Anurag Saha
Mar 12 at 9:04
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
$begingroup$
I'm glad to know that.
$endgroup$
– José Carlos Santos
Mar 12 at 9:47
add a comment |
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$begingroup$
$fracdy1+y=xdx$, you re-arranged wrongly.
$endgroup$
– Paul
Mar 12 at 8:26
$begingroup$
sorry, changed now
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– Anurag Saha
Mar 12 at 8:29
1
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What is the solution in the textbook? Maybe you have an equivalent expresssion.
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– Paras Khosla
Mar 12 at 8:38