Differential Equation

Tags

Calculus

Cegep/2

Word count

639 words

Reading time

4 minutes

Equation involving a function and some of its derivatives

[!abstract] Particular solution
Function $f$ that satisfies the DE when $f$ and its derivatives are substituted into the DE

[!abstract] General solution
General antiderivative of the particular solution

[!abstract]+ Separable differential equation
DE in which it is possible to factor the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ as a function of $x(h(x))$ times a function of $y(g(y))$

$$\frac{\mathrm{d}y}{\mathrm{d}x}=h(x)g(x)$$

[!example]+

• $y^{″}+3y=5y^{\prime }$
• $x^{″}(t)=-{\omega }^{2}x(t)$ (simple harmonic motion)
• $\frac{\mathrm{d}P}{\mathrm{d}t}=kP(t)$ (exponential growth)

Examples

Show that the function $y=5e^{2x}+3e^x$ is a solution to the DE $y^{″}-3y^{\prime }=-2y$.

$$\begin{array}{rl}y& =5e^{2x}+3e^x\\ y^{\prime }& =10e^{2x}+3e^x\\ y^{″}& =20e^{2x}+3e^x\\ \\ y^{″}-3y^{\prime }& =20e^{2x}+3e^x-3(10e^{2x}+3e^x)\\ & =20e^{2x}+3e^x-30e^{2x}-9e^x\\ & =-10e^{2x}-6e^x\\ & =-2(5e^{2x}+3e^x)\\ & =-2y\end{array}$$

Therefore, $y$ is a solution to the DE.

Consider $\frac{\mathrm{d}y}{\mathrm{d}x}=(2x-1)(1)$. Find the function $y$ that passes through the point $P(1,0)$.

$$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =(2x-1)(1)\\ dy& =(2x-1)dx\\ \int dy& =\int (2x-1)\mathrm{d}x\\ y& =x^2-x+C\end{array}$$

Substituting the coordinates into the general solution gives

$$\begin{array}{rl}0& =1^2-1+C\\ C& =0\\ \therefore y& =x^2-x\end{array}$$

Consider $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{x^2}{y^2}$. Find the function $y$ that passes through the point $P(1,0)$.

$$\begin{array}{rl}\frac{\mathrm{d}y}{\mathrm{d}x}& =\frac{x^2}{y^2}\\ y^{2}dy& =x^{2}dx\\ \int y^2\mathrm{d}y& =\int x^2\mathrm{d}x\\ \frac{1}{3}{y}^3& =\frac{1}{3}{x}^3+C\\ y^3& =x^3+3C\\ y& =\sqrt[3]{x^3+3C}\end{array}$$

Substituting the coordinates into the general solution gives

$$\begin{array}{rl}2& =\sqrt[3]{0^3+3C}\\ C& =\frac{8}{3}\\ \therefore y& =\sqrt[3]{x^3+8}\end{array}$$

Find the particular solution to $\frac{y^{\prime }}{x^2}=y$ at $(3,1)$.

$$\begin{array}{rl}\frac{y^{\prime }}{x^2}& =y\\ \frac{\frac{\mathrm{d}y}{\mathrm{d}x}}{x^2}& =y\\ \frac{\mathrm{d}y}{\mathrm{d}x}& =x^{2}y\\ \int \frac{1}{y}\mathrm{d}y& =\int x^2\mathrm{d}x\\ \ln |y|& =\frac{x^3}{3}+C\\ ⟹e^{\ln |y|}& =e^{\frac{x^3}{3}+C}\\ |y|& =e^{\frac{x^3}{3}+C}\\ y& =\pm e^{\frac{x^3}{3}+C}\end{array}$$

Substituting the coordinates gives

$$\begin{array}{rl}\ln |1|& =\frac{3^3}{3}+C\\ C& =-9\\ \therefore y& =\pm e^{\frac{x^3}{3}-9}\end{array}$$

Verifying the positive solution:

$$\begin{array}{rl}1& =e^{\frac{3^3}{3}-9}\\ 1& =e^0\\ 1& =1\end{array}$$

Verifying the negative solution:

$$\begin{array}{rl}1& =-e^{\frac{3^3}{3}-9}\\ 1& =-e^0\\ 1& =-1\end{array}$$

Therefore, the particular solution is $y=e^{\frac{x^3}{3}-9}$.

Solve $\frac{\mathrm{d}p}{\mathrm{d}t}=t^{2}p-p+t^2-1$.

$$\begin{array}{rl}\frac{\mathrm{d}p}{\mathrm{d}t}& =t^{2}p-p+t^2-1\\ & =p(t^2-1)+t^2-1\\ & =(p+1)(t^2-1)\\ \int \frac{1}{p+1}\mathrm{d}p& =\int (t^2-1)\mathrm{d}t\\ \ln |p+1|& =\frac{t^3}{3}-t+C\\ |p+1|& =e^{\frac{t^3}{3}-t+C}\\ p& =\pm e^{\frac{t^3}{3}-t+C}-1\end{array}$$

Solve $\frac{\mathrm{d}z}{\mathrm{d}t}+e^{t+z}=0$.

$$\begin{array}{rl}\frac{\mathrm{d}z}{\mathrm{d}t}+e^{t+z}& =0\\ \frac{\mathrm{d}z}{\mathrm{d}t}& =-e^t{e}^z\\ \int e^{-z}\mathrm{d}z& =-\int e^t\mathrm{d}t\\ -e^{-z}& =-e^t+C\\ z& =-\ln (e^t-C)\end{array}$$

Contributors

Ajitani Hifumi

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