Exact Method

When a prime indicates differentiation, the independent variable will be named if it is unambiguous; otherwise, a variable name must be specified. In the equations y′ = y, y′ = sin x and y′ = sin x + t, the independent variable is ambiguous and a dialog box appears asking for the independent variable. When a notation is used for differentiation that names the independent variable, it is taken from context and no dialog box appears.


$\blacktriangleright$ Solve ODE + Exact

y′ = sin x (Specify x), Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - cos x + C1

y′ = sin x (Specify t), Exact solution is: y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = $\left(\vphantom{ \sin x}\right.$sin x$\left.\vphantom{ \sin x}\right)$t + C1

y′ = sin x + t (Specify x), Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - cos x + tx + C1

y′ = sin x + t (Specify t), Exact solution is: y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = $\left(\vphantom{ \sin x}\right.$sin x$\left.\vphantom{ \sin x}\right)$t + ${\frac{{1}}{{2}}}$t2 + C1

y′ = y, Exact solution is: y$\left(\vphantom{ t}\right.$t$\left.\vphantom{ t}\right)$ = etC1


The constant C1 indicates that there is a family of solutions, one for each choice of C1. The following figure shows solutions for y′ = y corresponding to the choices 1/2, 1, 2, 3, and 4 for C1. To replicate this plot, drag solutions to the frame one at a time and then edit the colors on the Plot Components page of the Plot Properties dialog.

dtbpF3in2.0003in0pt

Scientific Notebook recognizes a variety of notation for a differential equation. The following examples illustrate some of this variety. The Leibniz notation ${\frac{{dy}}{{dx}}}$ or the Dx notation provides Scientific Notebook with enough information to determine the independent variable without a dialog box.


$\blacktriangleright$ Solve ODE + Exact

${\dfrac{{dy}}{{dx}}}$ = y + x, Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - x - 1 + exC1

y′ + xy = x (Specify x), Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = 1 + e-$\scriptstyle {\frac{{1}}{{2}}}$x2C


Dxy - y = sin x, Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - ${\frac{{1}}{{2}}}$cos x - ${\frac{{1}}{{2}}}$sin x + exC1


Following is a plot of three particular solutions for Dxy - y = sin x corresponding to C1 = 1, 2, 3. To replicate this plot, drag solutions to the frame one at a time and then edit the colors on the Plot Components page of the Plot Properties dialog.


$\blacktriangleright$ Plot 2D + Rectangular

- ${\frac{{1}}{{2}}}$cos x - ${\frac{{1}}{{2}}}$sin x + ex, - ${\frac{{1}}{{2}}}$cos x - ${\frac{{1}}{{2}% }}$sin x + 2ex, - ${\frac{{1}}{{2}}}$cos x - ${\frac{{1}}{{2}}}$sin x + 3exdtbpF3in2.0003in0pt

The three solutions can be distinguished by evaluation at 0. For example, the solution with C1 = 1 crosses the y-axis at y = 1/2.


$\blacktriangleright$ Solve ODE + Exact

y′′ + y = x2 (Specify x), Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - 2 + x2 + C1cos x + C2sin x


The following plot shows three solutions generated with constants $\left(\vphantom{ C_{1},C_{2}}\right.$C1, C2$\left.\vphantom{ C_{1},C_{2}}\right)$ = $\left(\vphantom{ 1,1}\right.$1, 1$\left.\vphantom{ 1,1}\right)$, $\left(\vphantom{ C_{1},C_{2}}\right.$C1, C2$\left.\vphantom{ C_{1},C_{2}}\right)$ = $\left(\vphantom{ 5,1}\right.$5, 1$\left.\vphantom{ 5,1}\right)$, and $\left(\vphantom{ C_{1},C_{2}}\right.$C1, C2$\left.\vphantom{ C_{1},C_{2}}\right)$ = $\left(\vphantom{ 1,5}\right.$1, 5$\left.\vphantom{ 1,5}\right)$. To replicate this plot, drag solutions to the frame one at a time and then edit the colors on the Plot Components page of the Plot Properties dialog.


$\blacktriangleright$ Plot 2D + Rectangular

x2 -2 + sin x + cos x, x2 -2 + 5 sin x + cos x, x2 -2 + sin x + 5 cos x

dtbpF3in2.0003in0pt

The particular solution y(x) = x2 -2 + sin x + 5 cos x is the one whose graph crosses the y-axis at y = 3.


$\blacktriangleright$ Solve ODE + Exact

${\dfrac{{d^{2}y}}{{dx^{2}}}}$ + ${\dfrac{{dy}}{{dx}}}$ = x + y, Exact solution is: y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = - 1 - x + C1e$\scriptstyle {\frac{{1}}{{2}}}$$\scriptstyle \left(\vphantom{ -1+\sqrt{5}}\right.$ -1 + $\scriptstyle \sqrt{{5}}$$\scriptstyle \left.\vphantom{ -1+\sqrt{5}}\right)$x + C2e-$\scriptstyle {\frac{{1}}{{2}}}$$\scriptstyle \left(\vphantom{ 1+\sqrt{5}}\right.$1 + $\scriptstyle \sqrt{{5}}$$\scriptstyle \left.\vphantom{ 1+\sqrt{5}}\right)$x


xy′ - y = x2, Exact solution is : y$\left(\vphantom{ x}\right.$x$\left.\vphantom{ x}\right)$ = x2 + xC1


Some differential equations that are not readily solvable by this method can be solved after rewriting the equation, as shown in the following two examples.



\begin{example} The differential equation $\frac{dy}{dx}=\frac{1}{x^{2}\sin y-x... ...^{2}}\sin y\,dy+C_{1}\right) e^{\frac{1}{2}y^{2}}$ \end{itemize} \end{example}


\begin{example} The differential equation $\frac{dy}{dx}+\frac{xy}{1-x^{2}}=x\s... ...ft( x\right) }=\allowbreak \frac{1}{3}\left( -1+x^{2}\right) x$. \end{example}