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Solving the linear first order differential equaition?

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

2

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
  $endgroup$
  – Fakemistake
  3 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|cite|improve this question

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why don't you separate the variables like $$frac{dy}{y}=-xdx?$$ Also the final conclusion is definitely wrong.
  $endgroup$
  – Fakemistake
  3 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

share|cite|improve this question

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

$endgroup$

share|cite|improve this question

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

share|cite|improve this question

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

edited 3 hours ago

Paras Khosla

951215

edited 3 hours ago

Paras Khosla

951215

asked 4 hours ago

A.SmithA.Smith

111

asked 4 hours ago

A.SmithA.Smith

111

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Paras KhoslaParas Khosla

951215

$endgroup$

add a comment | 

1

$begingroup$

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

$$y(x)=exp(-x^2/2).$$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

2217

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.

$endgroup$

add a comment | 

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2

$begingroup$

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.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Paras KhoslaParas Khosla

951215

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Paras KhoslaParas Khosla

951215

$endgroup$

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.

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Paras KhoslaParas Khosla

951215

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 4 hours ago

Paras KhoslaParas Khosla

951215

answered 4 hours ago

Paras KhoslaParas Khosla

951215

answered 4 hours ago

Paras KhoslaParas Khosla

951215

1

$begingroup$

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

$$y(x)=exp(-x^2/2).$$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

2217

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.

$endgroup$

add a comment | 

1

$begingroup$

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

$$y(x)=exp(-x^2/2).$$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

2217

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.

$endgroup$

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

$$y(x)=exp(-x^2/2).$$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

2217

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.

$endgroup$

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

st.mathst.math

2217

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 3 hours ago

st.mathst.math

2217

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 3 hours ago

st.mathst.math

2217