Thread: Substitution method to solve Differential equation

1. Substitution method to solve Differential equation

Hello,
How to solve this differential equation $y'=\frac{1}{XY(X^2\sin{Y^2}+1)}$?

2. Re: Substitution method to solve Differential equation

Well I am not sure if there is a specific reason you use capital letters on the right hand side, however, if my assertion that $\displaystyle y'(x) = \frac{\mathrm{d}Y(x)}{\mathrm{d}x}$, then the solution to the above can be found as follows:

Rewrite the equation using the relation $\displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}y} = 1$ which can be written as $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$. (Note: x=x(y) and y= y(x))

$\displaystyle \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} = \frac{1}{y(\sin(y^{2})x^2 + 1)x}$

Getting rid of the fraction (raising to the power of -1 so to speak) gives you:

$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}y} = y \sin(y^2)x^3+yx$

Moving $\displaystyle yx$ to the left hand side and diving both sides with $\displaystyle -\frac{1}{2}x^3$ you end up with:

$\displaystyle - \frac{2 \frac{\mathrm{d}x}{\mathrm{d}y}}{x^3} + \frac{2y}{x^2} = -2y\sin(y^2)$:

Defining $\displaystyle u(y) = \frac{1}{x^2}$ then $\displaystyle \frac{\mathrm{d}u}{\mathrm{d} y} = -\frac{2\frac{\mathrm{d}x}{\mathrm{d}y}}{x^3}$

This simplifies the equation to:

$\displaystyle \frac{\mathrm{d}u}{\mathrm{d}y} + 2yu=-2y \sin(y^2)$

Multiplying both sides with $\displaystyle e^{y^2}$ and substituting $\displaystyle 2e^{y^2} y = \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right)$ you find:

$\displaystyle e^{y^2} \frac{\mathrm{d}u}{\mathrm{d}y} + \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right) u(y) = -2e^{y^2} y \sin(y^2)$

Integrating both sides with respect to $\displaystyle y$ yields:

$\displaystyle e^{y^2} u = -\frac{1}{2}e^{y^2} \left( - \cos(y^2) + \sin(y^2)\right) + c$

Where $\displaystyle c$ is an arbitrary constant. Eliminate $\displaystyle e^{y^2}$ on both sides and you get:

$\displaystyle u(y) = \frac{1}{2}\left( \cos(y^2) - \sin(y^2) + 2e^{-y^2} c \right)$

Substitute back in the definition of $\displaystyle u(y)$ and solve for $\displaystyle y$ to get your solution

3. Re: Substitution method to solve Differential equation

Originally Posted by Krisly
Well I am not sure if there is a specific reason you use capital letters on the right hand side, however, if my assertion that $\displaystyle y'(x) = \frac{\mathrm{d}Y(x)}{\mathrm{d}x}$, then the solution to the above can be found as follows:

Rewrite the equation using the relation $\displaystyle \frac{\mathrm{d} y}{\mathrm{d}x} \frac{\mathrm{d}x}{\mathrm{d}y} = 1$ which can be written as $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}}$. (Note: x=x(y) and y= y(x))

$\displaystyle \frac{1}{\frac{\mathrm{d}x}{\mathrm{d}y}} = \frac{1}{y(\sin(y^{2})x^2 + 1)x}$

Getting rid of the fraction (raising to the power of -1 so to speak) gives you:

$\displaystyle \frac{\mathrm{d}x}{\mathrm{d}y} = y \sin(y^2)x^3+yx$

Moving $\displaystyle yx$ to the left hand side and diving both sides with $\displaystyle -\frac{1}{2}x^3$ you end up with:

$\displaystyle - \frac{2 \frac{\mathrm{d}x}{\mathrm{d}y}}{x^3} + \frac{2y}{x^2} = -2y\sin(y^2)$:

Defining $\displaystyle u(y) = \frac{1}{x^2}$ then $\displaystyle \frac{\mathrm{d}u}{\mathrm{d} y} = -\frac{2\frac{\mathrm{d}x}{\mathrm{d}y}}{x^3}$

This simplifies the equation to:

$\displaystyle \frac{\mathrm{d}u}{\mathrm{d}y} + 2yu=-2y \sin(y^2)$

Multiplying both sides with $\displaystyle e^{y^2}$ and substituting $\displaystyle 2e^{y^2} y = \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right)$ you find:

$\displaystyle e^{y^2} \frac{\mathrm{d}u}{\mathrm{d}y} + \frac{\mathrm{d}}{\mathrm{d}y} \left( e^{y^2} \right) u(y) = -2e^{y^2} y \sin(y^2)$

Integrating both sides with respect to $\displaystyle y$ yields:

$\displaystyle e^{y^2} u = -\frac{1}{2}e^{y^2} \left( - \cos(y^2) + \sin(y^2)\right) + c$

Where $\displaystyle c$ is an arbitrary constant. Eliminate $\displaystyle e^{y^2}$ on both sides and you get:

$\displaystyle u(y) = \frac{1}{2}\left( \cos(y^2) - \sin(y^2) + 2e^{-y^2} c \right)$

Substitute back in the definition of $\displaystyle u(y)$ and solve for $\displaystyle y$ to get your solution
Hello,
I want to know how to substitute u(y)? We have defined u(y)= $\frac{1}{x^2}$.Would you show me how to find y?

4. Re: Substitution method to solve Differential equation

Well replace $\displaystyle u(y)$ with $\displaystyle \frac{1}{x(y)^2}$ and isolate all terms of $\displaystyle y$ on the left-hand side and you should obtain:

$\displaystyle \frac{e^{y(x)^2}}{8 x^2} + \frac{1}{16 }e^{y(x)^2} (\sin(y(x)^2) - \cos(y(x)^2)) = c$

Also, as a sanity check, if you differentiate the above and solve for $\displaystyle y'$ you should find the expression you provided to begin with.

5. Re: Substitution method to solve Differential equation

Originally Posted by Krisly
Well replace $\displaystyle u(y)$ with $\displaystyle \frac{1}{x(y)^2}$ and isolate all terms of $\displaystyle y$ on the left-hand side and you should obtain:

$\displaystyle \frac{e^{y(x)^2}}{8 x^2} + \frac{1}{16 }e^{y(x)^2} (\sin(y(x)^2) - \cos(y(x)^2)) = c$

Also, as a sanity check, if you differentiate the above and solve for $\displaystyle y'$ you should find the expression you provided to begin with.
Hello,
As per my calculations i got $c=\frac{e^{y(x)^2}}{x^2}+\frac{e^{y(x)^2}}{2}( \sin y(x)^2-\cos y(x)^2)$. Where did i make a mistake? And meanwhile we have obtained c and not y. Now how to find y? Please explain me?

6. Re: Substitution method to solve Differential equation

Have you tried to differentiate your expression and solve for $\displaystyle y'$? Because when you do so, you find that the right-hand side evaluates to zero, as $\displaystyle c$ is a constant, hence multiplying your expression by 1/8 or mine with 8 doesn't change anything, meaning your expression are correct.

Regarding:
And meanwhile we have obtained c and not y. Now how to find y? Please explain me?
I can't seem to solve for $\displaystyle y(x)$ explicitly, this is the best I can provide you with. However, the equation we have found solves the differential equation.

Please also check that we are working on the same equation I am considering:

$\displaystyle y = \frac{1}{x \cdot y\left((x^2 \cdot \sin \left(y^2 \right)+1 \right)}$

the key is $\displaystyle \sin(y^2)$ instead of $\displaystyle \sin^2(y)$