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Solving the linear first order differential equation?
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Solving the linear first order differential equation?
First-order linear differential equationSolving a simple first order differential equationSolving a first order linear ODE via the method of integrating factorssolving second order linear differential equationIssue in first order differential equationDifferential Equation (First order with separable variable)First Order Differential Equation with Initial Value (Doesn't know how to remove the absolute sign)first order differential confusionHaving trouble exact first-order differential equation.first order differential equation and summation problem
$begingroup$
I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$frac{dy}{dx} = -xy $$
$$dy = -xy dx $$
$$int {dy} = int {-xy dx} $$
$$y = -frac{x^2}{2}y + c $$
$$y + frac{x^2}{2}y = c $$
$$y(1 + frac{x^2}{2}) = c $$
$$y = frac{c}{1 + frac{x^2}{2}} $$
I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.
I just know that what I did is wrong. Where did I go wrong, please? Thanks!
ordinary-differential-equations derivatives
edited 4 mins ago
David Richerby
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
$endgroup$
add a comment |
$begingroup$
I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$frac{dy}{dx} = -xy $$
$$dy = -xy dx $$
$$int {dy} = int {-xy dx} $$
$$y = -frac{x^2}{2}y + c $$
$$y + frac{x^2}{2}y = c $$
$$y(1 + frac{x^2}{2}) = c $$
$$y = frac{c}{1 + frac{x^2}{2}} $$
I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.
I just know that what I did is wrong. Where did I go wrong, please? Thanks!
ordinary-differential-equations derivatives
edited 4 mins ago
David Richerby
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
$endgroup$
$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago
add a comment |
$begingroup$
I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$frac{dy}{dx} = -xy $$
$$dy = -xy dx $$
$$int {dy} = int {-xy dx} $$
$$y = -frac{x^2}{2}y + c $$
$$y + frac{x^2}{2}y = c $$
$$y(1 + frac{x^2}{2}) = c $$
$$y = frac{c}{1 + frac{x^2}{2}} $$
I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.
I just know that what I did is wrong. Where did I go wrong, please? Thanks!
ordinary-differential-equations derivatives
edited 4 mins ago
David Richerby
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
$endgroup$
I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$frac{dy}{dx} = -xy $$
$$dy = -xy dx $$
$$int {dy} = int {-xy dx} $$
$$y = -frac{x^2}{2}y + c $$
$$y + frac{x^2}{2}y = c $$
$$y(1 + frac{x^2}{2}) = c $$
$$y = frac{c}{1 + frac{x^2}{2}} $$
I tried plugging in $0$ and $1$ respectively to find $c$, and got $c = 0$.
I just know that what I did is wrong. Where did I go wrong, please? Thanks!
ordinary-differential-equations derivatives
ordinary-differential-equations derivatives
edited 4 mins ago
David Richerby
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
edited 4 mins ago
David Richerby
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
edited 4 mins ago
David Richerby
2,18011324
edited 4 mins ago
David Richerby
2,18011324
edited 4 mins ago
David Richerby
2,18011324
2,18011324
asked 10 hours ago
A.SmithA.Smith
212
asked 10 hours ago
A.SmithA.Smith
212
asked 10 hours ago
A.SmithA.Smith
212
212
$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago
add a comment |
$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago
$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago
$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint:
This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfrac{mathrm dy}{mathrm dx}=-xy implies int dfrac{1}{y}mathrm dy =-int x mathrm dx$$
Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?
Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
$endgroup$
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
add a comment |
$begingroup$
After writing
$$intfrac{mathrm{d}y}{y}=int -x,mathrm{d}x,$$
you get $ln|y|=-frac{x^2}{2}+c_1$, which implies
$$y(x)=cexp(-x^2/2)quadtext{for}quad c=pmexp(c_1).$$
After plugging in $y(0)=1$, you get $c=1$, so
$$y(x)=exp(-x^2/2).$$
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
add a comment |
$begingroup$
Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^{int x, dx}=e^{x^2/2}$ to get that
$$
0=y'e^{x^2/2}+xye^{x^2/2}=frac{d}{dx}(ye^{x^2/2}).
$$
So
$$
ye^{x^2/2}=cimplies y=ce^{-x^2/2}
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfrac{mathrm dy}{mathrm dx}=-xy implies int dfrac{1}{y}mathrm dy =-int x mathrm dx$$
Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?
Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
$endgroup$
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
add a comment |
$begingroup$
Hint:
This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfrac{mathrm dy}{mathrm dx}=-xy implies int dfrac{1}{y}mathrm dy =-int x mathrm dx$$
Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?
Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
$endgroup$
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
add a comment |
$begingroup$
Hint:
This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfrac{mathrm dy}{mathrm dx}=-xy implies int dfrac{1}{y}mathrm dy =-int x mathrm dx$$
Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?
Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
$endgroup$
Hint:
This is a variable separable differential equation that is we can separate the differential equation such that for some functions $f(x)$ and $g(y)$: $f(x)mathrm dx =g(y)mathrm dy$. $$dfrac{mathrm dy}{mathrm dx}=-xy implies int dfrac{1}{y}mathrm dy =-int x mathrm dx$$
Now all that is left is to compute the integral and use the inital condition to figure out the value of constant. Can you proceed?
Note that you cannot simply integrate $-xy$ wrt $x$. Your computation assumes that $y$ is a constant which it most certainly is not.
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
edited 9 hours ago
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
answered 9 hours ago
Paras KhoslaParas Khosla
1,011215
1,011215
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
add a comment |
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
That helped a lot! Thank you!
$endgroup$
– A.Smith
3 hours ago
add a comment |
$begingroup$
After writing
$$intfrac{mathrm{d}y}{y}=int -x,mathrm{d}x,$$
you get $ln|y|=-frac{x^2}{2}+c_1$, which implies
$$y(x)=cexp(-x^2/2)quadtext{for}quad c=pmexp(c_1).$$
After plugging in $y(0)=1$, you get $c=1$, so
$$y(x)=exp(-x^2/2).$$
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
add a comment |
$begingroup$
After writing
$$intfrac{mathrm{d}y}{y}=int -x,mathrm{d}x,$$
you get $ln|y|=-frac{x^2}{2}+c_1$, which implies
$$y(x)=cexp(-x^2/2)quadtext{for}quad c=pmexp(c_1).$$
After plugging in $y(0)=1$, you get $c=1$, so
$$y(x)=exp(-x^2/2).$$
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
add a comment |
$begingroup$
After writing
$$intfrac{mathrm{d}y}{y}=int -x,mathrm{d}x,$$
you get $ln|y|=-frac{x^2}{2}+c_1$, which implies
$$y(x)=cexp(-x^2/2)quadtext{for}quad c=pmexp(c_1).$$
After plugging in $y(0)=1$, you get $c=1$, so
$$y(x)=exp(-x^2/2).$$
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
After writing
$$intfrac{mathrm{d}y}{y}=int -x,mathrm{d}x,$$
you get $ln|y|=-frac{x^2}{2}+c_1$, which implies
$$y(x)=cexp(-x^2/2)quadtext{for}quad c=pmexp(c_1).$$
After plugging in $y(0)=1$, you get $c=1$, so
$$y(x)=exp(-x^2/2).$$
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 8 hours ago
answered 8 hours ago
st.mathst.math
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 8 hours ago
st.mathst.math
2807
answered 8 hours ago
st.mathst.math
2807
2807
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
st.math is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
add a comment |
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
Thank you so so much! I super appreciate it!
$endgroup$
– A.Smith
3 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
$begingroup$
You're welcome :)
$endgroup$
– st.math
2 hours ago
add a comment |
$begingroup$
Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^{int x, dx}=e^{x^2/2}$ to get that
$$
0=y'e^{x^2/2}+xye^{x^2/2}=frac{d}{dx}(ye^{x^2/2}).
$$
So
$$
ye^{x^2/2}=cimplies y=ce^{-x^2/2}
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^{int x, dx}=e^{x^2/2}$ to get that
$$
0=y'e^{x^2/2}+xye^{x^2/2}=frac{d}{dx}(ye^{x^2/2}).
$$
So
$$
ye^{x^2/2}=cimplies y=ce^{-x^2/2}
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
add a comment |
$begingroup$
Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^{int x, dx}=e^{x^2/2}$ to get that
$$
0=y'e^{x^2/2}+xye^{x^2/2}=frac{d}{dx}(ye^{x^2/2}).
$$
So
$$
ye^{x^2/2}=cimplies y=ce^{-x^2/2}
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
$endgroup$
Here is another method that does not use separation of variables and uses integrating factors. Write
$$
y'+xy=0
$$
and multiply both sides by $e^{int x, dx}=e^{x^2/2}$ to get that
$$
0=y'e^{x^2/2}+xye^{x^2/2}=frac{d}{dx}(ye^{x^2/2}).
$$
So
$$
ye^{x^2/2}=cimplies y=ce^{-x^2/2}
$$
for some $c$. Since $y(0)=1$, we deduce that
$$
y=exp(-x^2/2).
$$
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
answered 2 hours ago
Foobaz JohnFoobaz John
22.2k41452
22.2k41452
add a comment |
add a comment |
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$begingroup$
Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
$endgroup$
– Fakemistake
9 hours ago