How To Solve Differential Equations

Below are a few examples of ordinary differential equations. dydx=ky{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=ky} d2xdt2+kx=0{\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+kx=0} Below are a few examples of partial differential equations. ∂2f∂x2+∂2f∂y2=0{\displaystyle {\frac {\partial ^{2}f}{\partial x^{2}}}+{\frac {\partial ^{2}f}{\partial y^{2}}}=0} ∂u∂t−α∂2u∂x2=0{\displaystyle {\frac {\partial u}{\partial t}}-\alpha {\frac {\partial ^{2}u}{\partial x^{2}}}=0}

For example, the equation below is a third-order, second degree equation. (d3ydx3)2+dydx=0{\displaystyle \left({\frac {\mathrm {d} ^{3}y}{\mathrm {d} x^{3}}}\right)^{2}+{\frac {\mathrm {d} y}{\mathrm {d} x}}=0}

Below are a few examples of linear differential equations. dydx+p(x)y=q(x){\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}+p(x)y=q(x)} x2d2ydx2+axdydx+by=0{\displaystyle x^{2}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+ax{\frac {\mathrm {d} y}{\mathrm {d} x}}+by=0} Below are a few examples of nonlinear differential equations. The first equation is nonlinear because of the sine term. d2θdt2+glsin⁡θ=0{\displaystyle {\frac {\mathrm {d} ^{2}\theta }{\mathrm {d} t^{2}}}+{\frac {g}{l}}\sin \theta =0} d2xdt2+(dxdt)2+tx2=0{\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+\left({\frac {\mathrm {d} x}{\mathrm {d} t}}\right)^{2}+tx^{2}=0}

For example, the equation below is one that we will discuss how to solve in this article. It is a second-order linear differential equation. Its general solution contains two arbitrary constants. To evaluate these constants, we also require initial conditions at x(0){\displaystyle x(0)} and x′(0). {\displaystyle x’(0). } These initial conditions are usually given at x=0,{\displaystyle x=0,} but they do not have to be. We will also discuss finding particular solutions given initial conditions later in the article. d2xdt2+k2x=0{\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+k^{2}x=0} x(t)=c1cos⁡kt+c2sin⁡kt{\displaystyle x(t)=c_{1}\cos kt+c_{2}\sin kt}

p(x)=0.{\displaystyle p(x)=0.} By the fundamental theorem of calculus, the integral of a derivative of a function is the function itself. We can then simply integrate to obtain our answer. Remember that evaluating an indefinite integral introduces an arbitrary constant.

dydx−2ysin⁡x={\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}-2y\sin x=0} 12ydy=sin⁡xdx12ln⁡y=−cos⁡x+Cln⁡y=−2cos⁡x+Cy(x)=Ce−2cos⁡x{\displaystyle {\begin{aligned}{\frac {1}{2y}}\mathrm {d} y&=\sin x\mathrm {d} x\{\frac {1}{2}}\ln y&=-\cos x+C\\ln y&=-2\cos x+C\y(x)&=Ce^{-2\cos x}\end{aligned}}}

tdydt+2y=t2,y(2)=3{\displaystyle t{\frac {\mathrm {d} y}{\mathrm {d} t}}+2y=t^{2},\quad y(2)=3} dydt+2ty=t{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} t}}+{\frac {2}{t}}y=t} μ(t)=e∫p(t)dt=e2ln⁡t=t2{\displaystyle \mu (t)=e^{\int p(t)\mathrm {d} t}=e^{2\ln t}=t^{2}} ddt(t2y)=t3t2y=14t4+Cy(t)=14t2+Ct2{\displaystyle {\begin{aligned}{\frac {\mathrm {d} }{\mathrm {d} t}}(t^{2}y)&=t^{3}\t^{2}y&={\frac {1}{4}}t^{4}+C\y(t)&={\frac {1}{4}}t^{2}+{\frac {C}{t^{2}}}\end{aligned}}} 3=y(2)=1+C4,C=8{\displaystyle 3=y(2)=1+{\frac {C}{4}},\quad C=8} y(t)=14t2+8t2{\displaystyle y(t)={\frac {1}{4}}t^{2}+{\frac {8}{t^{2}}}}

∫dyh(y)=∫g(x)dx{\displaystyle \int {\frac {\mathrm {d} y}{h(y)}}=\int g(x)\mathrm {d} x} Example 1. 3. dydx=x3y(1+x4){\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {x^{3}}{y(1+x^{4})}}} ∫ydy=∫x31+x4dx12y2=14ln⁡(1+x4)+Cy(x)=12ln⁡(1+x4)+C{\displaystyle {\begin{aligned}\int y\mathrm {d} y&=\int {\frac {x^{3}}{1+x^{4}}}\mathrm {d} x\{\frac {1}{2}}y^{2}&={\frac {1}{4}}\ln(1+x^{4})+C\y(x)&={\frac {1}{2}}\ln(1+x^{4})+C\end{aligned}}}

dydx=g(x,y)h(x,y).{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {g(x,y)}{h(x,y)}}.} Let g(x,y){\displaystyle g(x,y)} and h(x,y){\displaystyle h(x,y)} be functions of x{\displaystyle x} and y.{\displaystyle y.} Then a homogeneous differential equation is an equation where g{\displaystyle g} and h{\displaystyle h} are homogeneous functions of the same degree. That is to say, the function satisfies the property g(αx,αy)=αkg(x,y),{\displaystyle g(\alpha x,\alpha y)=\alpha ^{k}g(x,y),} where k{\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 v=y/x{\displaystyle v=y/x} or v=x/y.{\displaystyle v=x/y.}

dydx=x3y(1+x4){\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {x^{3}}{y(1+x^{4})}}} ∫ydy=∫x31+x4dx12y2=14ln⁡(1+x4)+Cy(x)=12ln⁡(1+x4)+C{\displaystyle {\begin{aligned}\int y\mathrm {d} y&=\int {\frac {x^{3}}{1+x^{4}}}\mathrm {d} x\{\frac {1}{2}}y^{2}&={\frac {1}{4}}\ln(1+x^{4})+C\y(x)&={\frac {1}{2}}\ln(1+x^{4})+C\end{aligned}}}

dydx=y3−x3y2x{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {y^{3}-x^{3}}{y^{2}x}}} We first observe that this is a nonlinear equation in y. {\displaystyle y. } We also see that this equation cannot be separated. However, it is a homogeneous differential equation because both the top and bottom are homogeneous of degree 3. Therefore, we can make the change of variables v=y/x. {\displaystyle v=y/x. } dydx=yx−x2y2=v−1v2{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {y}{x}}-{\frac {x^{2}}{y^{2}}}=v-{\frac {1}{v^{2}}}} y=vx,dydx=dvdxx+v{\displaystyle y=vx,\quad {\frac {\mathrm {d} y}{\mathrm {d} x}}={\frac {\mathrm {d} v}{\mathrm {d} x}}x+v} dvdxx=−1v2. {\displaystyle {\frac {\mathrm {d} v}{\mathrm {d} x}}x=-{\frac {1}{v^{2}}}. } This is now a separable equation in v. {\displaystyle v. } v(x)=−3ln⁡x+C3{\displaystyle v(x)={\sqrt[{3}]{-3\ln x+C}}} y(x)=x−3ln⁡x+C3{\displaystyle y(x)=x{\sqrt[{3}]{-3\ln x+C}}}

∂M∂y=∂N∂x{\displaystyle {\frac {\partial M}{\partial y}}={\frac {\partial N}{\partial x}}}

3x2+y2+2xydydx=0{\displaystyle 3x^{2}+y^{2}+2xy{\frac {\mathrm {d} y}{\mathrm {d} x}}=0} φ=∫(3x2+y2)dx=x3+xy2+c(y)∂φ∂y=N(x,y)=2xy+dcdy{\displaystyle {\begin{aligned}\varphi &=\int (3x^{2}+y^{2})\mathrm {d} x=x^{3}+xy^{2}+c(y)\{\frac {\partial \varphi }{\partial y}}&=N(x,y)=2xy+{\frac {\mathrm {d} c}{\mathrm {d} y}}\end{aligned}}} dcdy=0,c(y)=C{\displaystyle {\frac {\mathrm {d} c}{\mathrm {d} y}}=0,\quad c(y)=C} x3+xy2=C{\displaystyle x^{3}+xy^{2}=C}

Characteristic equation. This differential equation is notable because we can solve it very easily if we make some observations about what properties its solutions must have. This equation tells us that y{\displaystyle y} and its derivatives are all proportional to each other. From our previous examples in dealing with first-order equations, we know that only the exponential function has this property. Therefore, we will put forth an ansatz – an educated guess – on what the solution will be.

W=|y1y2y1′y2′|{\displaystyle W={\begin{vmatrix}y_{1}&y_{2}\y_{1}’&y_{2}’\end{vmatrix}}}

eαx(c1cos⁡βx+ic1sin⁡βx+c2cos⁡βx−ic2sin⁡βx){\displaystyle e^{\alpha x}(c_{1}\cos \beta x+ic_{1}\sin \beta x+c_{2}\cos \beta x-ic_{2}\sin \beta x)}

d2xdt2+3dxdt+10x=0,x(0)=1, x′(0)=−1{\displaystyle {\frac {\mathrm {d} ^{2}x}{\mathrm {d} t^{2}}}+3{\frac {\mathrm {d} x}{\mathrm {d} t}}+10x=0,\quad x(0)=1,\ x’(0)=-1} r2+3r+10=0,r±=−3±9−402=−32±312i{\displaystyle r^{2}+3r+10=0,\quad r_{\pm }={\frac {-3\pm {\sqrt {9-40}}}{2}}=-{\frac {3}{2}}\pm {\frac {\sqrt {31}}{2}}i} x(t)=e−3t/2(c1cos⁡312t+c2sin⁡312t){\displaystyle x(t)=e^{-3t/2}\left(c_{1}\cos {\frac {\sqrt {31}}{2}}t+c_{2}\sin {\frac {\sqrt {31}}{2}}t\right)} x(0)=1=c1{\displaystyle x(0)=1=c_{1}} x′(t)=−32e−3t/2(c1cos⁡312t+c2sin⁡312t)+e−3t/2(−312c1sin⁡312t+312c2cos⁡312t){\displaystyle {\begin{aligned}x’(t)&=-{\frac {3}{2}}e^{-3t/2}\left(c_{1}\cos {\frac {\sqrt {31}}{2}}t+c_{2}\sin {\frac {\sqrt {31}}{2}}t\right)\&+e^{-3t/2}\left(-{\frac {\sqrt {31}}{2}}c_{1}\sin {\frac {\sqrt {31}}{2}}t+{\frac {\sqrt {31}}{2}}c_{2}\cos {\frac {\sqrt {31}}{2}}t\right)\end{aligned}}} x′(0)=−1=−32c1+312c2,c2=131{\displaystyle x’(0)=-1=-{\frac {3}{2}}c_{1}+{\frac {\sqrt {31}}{2}}c_{2},\quad c_{2}={\frac {1}{\sqrt {31}}}} x(t)=e−3t/2(cos⁡312t+131sin⁡312t){\displaystyle x(t)=e^{-3t/2}\left(\cos {\frac {\sqrt {31}}{2}}t+{\frac {1}{\sqrt {31}}}\sin {\frac {\sqrt {31}}{2}}t\right)}

Repeated roots to the homogeneous differential equation with constant coefficients. Recall that a second-order equation should have two linearly independent solutions. If the characteristic equation yields a repeating root, then the solution set fails to span the space because the solutions are linearly dependent. We must then use reduction of order to find the second linearly independent solution.

d2ydx2+8dydx+16y=0{\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+8{\frac {\mathrm {d} y}{\mathrm {d} x}}+16y=0} y=v(x)e−4xy′=v′(x)e−4x−4v(x)e−4xy″=v″(x)e−4x−8v′(x)e−4x+16v(x)e−4x{\displaystyle {\begin{aligned}y&=v(x)e^{-4x}\y’&=v’(x)e^{-4x}-4v(x)e^{-4x}\y’’&=v’’(x)e^{-4x}-8v’(x)e^{-4x}+16v(x)e^{-4x}\end{aligned}}} v″e−4x−8v′e−4x+16ve−4x+8v′e−4x−32ve−4x+16ve−4x=0{\displaystyle {\begin{aligned}v’’e^{-4x}&-{\cancel {8v’e^{-4x}}}+{\cancel {16ve^{-4x}}}\&+{\cancel {8v’e^{-4x}}}-{\cancel {32ve^{-4x}}}+{\cancel {16ve^{-4x}}}=0\end{aligned}}}

Characteristic equation. The structure of this differential equation is such that each term is multiplied by a power term whose degree is equal to the order of the derivative.

n±=1−a±(a−1)2−4b2{\displaystyle n_{\pm }={\frac {1-a\pm {\sqrt {(a-1)^{2}-4b}}}{2}}}

Method of undetermined coefficients. The method of undetermined coefficients is a method that works when the source term is some combination of exponential, trigonometric, hyperbolic, or power terms. These terms are the only terms that have a finitely many number of linearly independent derivatives. In this section, we concentrate on finding the particular solution. [8] X Research source

d2ydt2+6y=2e3t−cos⁡5t{\displaystyle {\frac {\mathrm {d} ^{2}y}{\mathrm {d} t^{2}}}+6y=2e^{3t}-\cos 5t} yc(t)=c1cos⁡6t+c2sin⁡6t{\displaystyle y_{c}(t)=c_{1}\cos {\sqrt {6}}t+c_{2}\sin {\sqrt {6}}t} yp(t)=Ae3t+Bcos⁡5t+Csin⁡5t{\displaystyle y_{p}(t)=Ae^{3t}+B\cos 5t+C\sin 5t} 9Ae3t−25Bcos⁡5t−25Csin⁡5t+6Ae3t+6Bcos⁡5t+6Csin⁡5t=2e3t−cos⁡5t{\displaystyle {\begin{aligned}9Ae^{3t}-25B\cos 5t&-25C\sin 5t+6Ae^{3t}\&+6B\cos 5t+6C\sin 5t=2e^{3t}-\cos 5t\end{aligned}}} {9A+6A=2,A=215−25B+6B=−1,B=119−25C+6C=0,C=0{\displaystyle {\begin{cases}9A+6A=2,&A={\dfrac {2}{15}}\-25B+6B=-1,&B={\dfrac {1}{19}}\-25C+6C=0,&C=0\end{cases}}} y(t)=c1cos⁡6t+c2sin⁡6t+215e3t+119cos⁡5t{\displaystyle y(t)=c_{1}\cos {\sqrt {6}}t+c_{2}\sin {\sqrt {6}}t+{\frac {2}{15}}e^{3t}+{\frac {1}{19}}\cos 5t}

v1′y1+v2′y2=0{\displaystyle v_{1}‘y_{1}+v_{2}‘y_{2}=0} y′=v1y1′+v2y2′{\displaystyle y’=v_{1}y_{1}’+v_{2}y_{2}’} y″=v1′y1′+v1y1″+v2′y2′+v2y2″{\displaystyle y’’=v_{1}‘y_{1}’+v_{1}y_{1}’’+v_{2}‘y_{2}’+v_{2}y_{2}’’}

dydx=kx{\displaystyle {\frac {\mathrm {d} y}{\mathrm {d} x}}=kx}

x¨+2βx˙+ω02x=F(t){\displaystyle {\ddot {x}}+2\beta {\dot {x}}+\omega _{0}^{2}x=F(t)}

x2d2ydx2+xdydx+(x2−α2)y=0{\displaystyle x^{2}{\frac {\mathrm {d} ^{2}y}{\mathrm {d} x^{2}}}+x{\frac {\mathrm {d} y}{\mathrm {d} x}}+(x^{2}-\alpha ^{2})y=0}

∇⋅E=ρϵ0∇⋅B=0∇×E=−∂B∂t∇×B=μ0J+μ0ϵ0∂E∂t{\displaystyle {\begin{aligned}\nabla \cdot \mathbf {E} &={\frac {\rho }{\epsilon _{0}}}\\nabla \cdot \mathbf {B} &=0\\nabla \times \mathbf {E} &=-{\frac {\partial \mathbf {B} }{\partial t}}\\nabla \times \mathbf {B} &=\mu _{0}\mathbf {J} +\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}\end{aligned}}}

iℏ∂Ψ∂t=H^Ψ{\displaystyle i\hbar {\frac {\partial \Psi }{\partial t}}={\hat {H}}\Psi } iℏ∂Ψ∂t=(−ℏ22m∇2+V(r,t))Ψ{\displaystyle i\hbar {\frac {\partial \Psi }{\partial t}}=\left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+V(\mathbf {r} ,t)\right)\Psi }

∂2u∂t2=c2∇2u{\displaystyle {\frac {\partial ^{2}u}{\partial t^{2}}}=c^{2}\nabla ^{2}u} u(x,t)=f(x−ct)+g(x+ct){\displaystyle u(x,t)=f(x-ct)+g(x+ct)}

∂u∂t+(u⋅∇)u−ν∇2u=−∇h,∂ρ∂t+∇⋅(ρu)=0{\displaystyle {\frac {\partial \mathbf {u} }{\partial t}}+(\mathbf {u} \cdot \nabla )\mathbf {u} -\nu \nabla ^{2}\mathbf {u} =-\nabla h,\quad {\frac {\partial \rho }{\partial t}}+\nabla \cdot (\rho \mathbf {u} )=0}