Practice for Solving Exact ODEs

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Why?

Solve the following exact equations, implicit general solutions will suffice:

$(2 xy + x^2) dx + (x^2+y^2+1) dy = 0$ , $\answer {x^2y + \frac {x^3}{3} + \frac {y^3}{3} + y} = C$
$x^5 + y^5 \frac {dy}{dx} = 0$ , $\answer {\frac {x^6}{6} + \frac {y^6}{6}} = C$
$e^x+y^3 + 3xy^2 \frac {dy}{dx} = 0$ , $\answer {e^x + xy^3} = C$
$(x+y)\cos (x)+\sin (x) + \sin (x)y' = 0$ , $\answer {(x+y)\sin (x) + \cos (x)} = C$

Solve the following exact equations, implicit general solutions will suffice:

$\cos (x)+ye^{xy} + xe^{xy} y' = 0$ , $\answer {e^{xy}+\sin (x)} = C$
$(2x+y)dx + (x-4y) dy = 0$ , $\answer {x^2+xy-2y^2} = C$
$e^x + e^y \frac {dy}{dx} = 0$ , $\answer {e^x+e^y} = C$
$(3x^2+3y)dx + (3y^2+3x)dy = 0$ , $\answer {x^3 + 3xy+ y^3} = C$

Solve the differential equation $(2ye^{2xy} - 2x) + (2xe^{2xy} + \cos (y))y' = 0$ $C = \answer {e^{2xy} - x^2 + \sin (y)}$

Solve the differential equation $(-y\sin (xy) - 2xe^{x^2}) + (-x\sin (xy) + 1)y' = 0$ $C = \answer {\cos (xy) - e^{x^2} + y}$

Solve the differential equation $(2x + 3y\sin (xy)) + (3x\sin (xy) - e^y)y' = 0$ with $y(2) = 0$ .

[Note: Make sure to enter the full equation, e.g. 0 = x + 4] $\answer {x^2 - 3\cos (xy) - e^y = 0}$

Solve the differential equation $x + yy' = 0$ with $y(0) = 8$ . Write this as an explicit function. [Note: Make sure to enter the full equation, e.g. 0 = x + 4] $\answer {y = \sqrt {64-x^2}}$

Determine the interval of $x$ values where the solution is valid: ([ $\answer {-8},\answer {8}$ )]

Solve the differential equation $2x-2 + (8y+16)y' = 0$ with $y(2) = 0$ . Write this as an explicit function. [Note: Make sure to enter the full equation, e.g. 0 = x + 4] $\answer {y = \frac {-4 + \sqrt {17 - (x-1)^2}}{2}}$

Determine the interval of $x$ values where the solution is valid: ([ $\answer {1 - \sqrt {17}},\answer {1 + \sqrt {17}}$ )]

Find the integrating factor for the following equations making them into exact equations. You can either use the formulas in this section or guess what the integrating factor should be.

$e^{xy} dx + \frac {y}{x} e^{xy} dy = 0$ , $\answer {xe^{-xy}}$
$\frac {e^x+y^3}{y^2} dx + 3x dy = 0$ , $\answer {y^2}$
$(4x^5 + 9x^4y^2 + 10\frac {y}{x})dx + (6x^5y + 3x^2y^2 - 5)dy = 0$ , $\answer {x^{-2}}$
$2\sin (y) dx + x\cos (y)dy = 0$ , $\answer {x}$

Find the integrating factor for the following equations making them into exact equations:

$\frac {1}{y}dx + 3y dy = 0$ , Integrating Factor: $\answer {y}$ , Exact Equation: $\answer {dx + 3y^2dy = 0}$
$dx - e^{-x-y} dy = 0$ , $\answer {xe^{-xy}}$ , Integrating Factor: $\answer {e^x}$ , Exact Equation: $\answer {e^x dx - e^{-y}dy= 0}$
$\bigl ( \frac {\cos (x)}{y^2} + \frac {1}{y} \bigr ) dx + \frac {x}{y^2} dy = 0$ , $\answer {xe^{-xy}}$ , Integrating Factor: $\answer {y^2}$ , Exact Equation: $\answer {(\cos (x)+y)dx + xdy = 0}$
$\bigl ( 2y + \frac {y^2}{x} \bigr ) dx + ( 2y+x )dy = 0$ , $\answer {xe^{-xy}}$ , Integrating Factor: $\answer {x}$ , Exact Equation: $\answer {( 2xy+y^2 )dx + (x^2+2xy)dy = 0}$

Suppose you have an equation of the form: $f(x) + g(y) \frac {dy}{dx} = 0$ .

Is it exact? Yes.No.

Find the form of the potential function in terms of $f$ and $g$ . $\int \answer {f(x)}dx + \int \answer {g(y)}dy = 0.$

Suppose that we have the equation $f(x) dx - dy = 0$ .

Is this equation exact? Yes.No.

Find the general solution using a definite integral. $y = \int \answer {f(x)}dx + C$

Find the potential function $F(x,y)$ of the exact equation $\frac {1+xy}{x}dx + \bigl (\frac {1}{y} + x \bigr ) dy = 0$ in two different ways.

Integrate $M$ in terms of $x$ and then differentiate in $y$ and set to $N$ . $\answer {xy + \ln (x) + \ln (y) = C}$
Integrate $N$ in terms of $y$ and then differentiate in $x$ and set to $M$ . $\answer {xy + \ln (x) + \ln (y) = C}$

A function $u(x,y)$ is said to be a harmonic function if $u_{xx} + u_{yy} = 0$ .

Compute: the difference between $M_y = -u_{yy}$ and $N_x = u_{xx}$ : $\answer {0}$ .

Since the difference is zero, they are equal, thus $-u_y dx + u_x dy = 0$ is an exact equation. So there exists (at least locally) the so-called harmonic conjugate function $v(x,y)$ such that $v_x = -u_y$ and $v_y = u_x$ .

Verify that the following $u$ are harmonic and find the corresponding harmonic conjugates $v$ :

$u = 2xy$ , corresponding harmonic conjugate: $\answer {y^2 - x^2}$ .
$u = e^x \cos y$ , corresponding harmonic conjugate: $\answer {e^{x}\sin (y)}$ .
$u = x^3-3xy^2$ , corresponding harmonic conjugate: $\answer {3x^2y - y^3}$ .

Consider equations of the form $y' = f(x)g(y)$ . Compute: $M_y = \answer {0}$ , and $N_x = \answer {0}$ .

Since $M_y = 0 = N_x$ , we know that it is an exact function. We can thus use the substitution $y' = xy$ to rewrite it as an exact equation and solve. Start by using separation of variables to get: $\answer {-x }dx + \answer {\frac {1}{y}} dy = 0$

Then taking the integral we get: $F(x,y) = \answer {-\frac {x^2}{2} + \ln \lvert y \rvert } + C$