Find a Differential Equation That Has Solution P T

Nonlinear first-order equations. In this section, we discuss the methods of solving certain nonlinear first-order differential equations. There is no general solution in closed form, but certain equations are able to be solved using the techniques below.^[3]

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=f(x,y)}$

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=h(x)g(y).}$ If the function ${\displaystyle f(x,y)=h(x)g(y)}$ can be separated into functions of one variable each, then the equation is said to be separable. We then proceed with the same method as before.

• ${\displaystyle \int {\frac {\mathrm {d} y}{h(y)}}=\int g(x)\mathrm {d} x}$
• Example 1.3.

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {g(x,y)}{h(x,y)}}.}$ Let ${\displaystyle g(x,y)}$ and ${\displaystyle h(x,y)}$ be functions of ${\displaystyle x}$ and ${\displaystyle y.}$ Then a homogeneous differential equation is an equation where ${\displaystyle g}$ and ${\displaystyle h}$ are homogeneous functions of the same degree. That is to say, the function satisfies the property ${\displaystyle g(\alpha x,\alpha y)=\alpha ^{k}g(x,y),}$ where ${\displaystyle k}$ is called the degree of homogeneity. Every homogeneous differential equation can be converted into a separable equation through a sufficient change of variables, either ${\displaystyle v=y/x}$ or ${\displaystyle v=x/y.}$

• Example 1.4. The above discussion regarding homogeneity may be somewhat arcane. Let us see how this applies through an example.

${\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=p(x)y+q(x)y^{n}.}$ This is the Bernoulli differential equation, a particular example of a nonlinear first-order equation with solutions that can be written in terms of elementary functions.

${\displaystyle M(x,y)+N(x,y){\frac {\mathrm {d} y}{\mathrm {d} x}}=0.}$ Here, we discuss exact equations. We wish to find a function ${\displaystyle \varphi (x,y),}$ called the potential function, such that ${\displaystyle {\frac {\mathrm {d} \varphi }{\mathrm {d} x}}=0.}$

• To fulfill this condition, we have the following total derivative. The total derivative allows for additional variable dependencies. To calculate the total derivative of ${\displaystyle \varphi }$ with respect to ${\displaystyle x,}$ we allow for the possibility that ${\displaystyle y}$ may also depend on ${\displaystyle x.}$
  • ${\displaystyle {\frac {\mathrm {d} \varphi }{\mathrm {d} x}}={\frac {\partial \varphi }{\partial x}}+{\frac {\partial \varphi }{\partial y}}{\frac {\mathrm {d} y}{\mathrm {d} x}}}$
• Comparing terms, we have ${\displaystyle M(x,y)={\frac {\partial \varphi }{\partial x}}}$ and ${\displaystyle N(x,y)={\frac {\partial \varphi }{\partial y}}.}$ It is a standard result from multivariable calculus that mixed derivatives for smooth functions are equal to each other. This is sometimes known as Clairaut's theorem. The differential equation is then exact if the following condition holds.
  • ${\displaystyle {\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}}$
• The method of solving exact equations is similar to finding potential functions in multivariable calculus, which we go into very shortly. We first integrate ${\displaystyle M}$ with respect to ${\displaystyle x.}$ Because ${\displaystyle M}$ is a function of both ${\displaystyle x}$ and ${\displaystyle y,}$ the integration can only partially recover ${\displaystyle \varphi ,}$ which the term ${\displaystyle {\tilde {\varphi }}}$ is intended to remind the reader of. There is also an integration constant that is a function of ${\displaystyle y.}$
  • ${\displaystyle \varphi (x,y)=\int M(x,y)\mathrm {d} x={\tilde {\varphi }}(x,y)+c(y)}$
• We then take the partial derivative of our result with respect to ${\displaystyle y,}$ compare terms with ${\displaystyle N(x,y),}$ and integrate to obtain ${\displaystyle c(y).}$ We can also start by integrating ${\displaystyle N}$ first and then taking the partial derivative of our result with respect to ${\displaystyle x}$ to solve for the arbitrary function ${\displaystyle d(x).}$ Either method is fine, and usually, the simpler function to integrate is chosen.
  • ${\displaystyle N(x,y)={\frac {\partial \varphi }{\partial y}}={\frac {\partial {\tilde {\varphi }}}{\partial y}}+{\frac {\mathrm {d} c}{\mathrm {d} y}}}$
• Example 1.5. We can check that the equation below is exact by doing the partial derivatives.
• If our differential equation is not exact, then there are certain instances where we can find an integrating factor that makes it exact. However, these equations are even harder to find applications of in the sciences, and integrating factors, though guaranteed to exist, are not at all guaranteed to easily be found. As such, we will not go into them here.

Find a Differential Equation That Has Solution P T

Source: https://www.wikihow.com/Solve-Differential-Equations

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