Basic Differential Equations and Solutions

A differential equation connects an unknown function with its derivatives. The order of the equation is the order of the highest derivative in it: ${y'=2x}$ is a first-order equation, and ${y''+y=0}$ is a second-order one. To solve a differential equation means to find every function that satisfies it.

The answer usually contains arbitrary constants: one for a first-order equation, two for a second-order equation. It is called the general solution. For example, the general solution of ${y'=2x}$ is ${y=x^2+c},$ a whole family of parabolas. An extra condition such as ${y(0)=2}$ picks one curve from the family, here ${c=2}.$

Below are the standard types of equations that can be solved by a formula. For each type you will find the form of the equation, the formula for its solution, the idea behind the formula and a worked example. Throughout the page $y'$ is the derivative ${dy/dx},$ and $c,$ $C,$ $c_1,$ $c_2$ are arbitrary constants.

On this page: Quick reference · Separable · Linear first-order · Bernoulli · Exact · Homogeneous · yF(xy)dx + xG(xy)dy = 0 · Second-order homogeneous · Second-order nonhomogeneous · Euler · Bessel · Legendre · Exercises

Quick Reference

Equation	How to solve it
${f_1(x)g_1(y)\,dx+f_2(x)g_2(y)\,dy=0}$	Separate the variables and integrate
${y'+P(x)\,y=Q(x)}$	Multiply by the integrating factor ${e^{\int P\,dx}}$
${y'+P(x)\,y=Q(x)\,y^n}$	Substitute ${v=y^{1-n}}$
${M\,dx+N\,dy=0},$ ${\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}$	Find $F$ with ${dF=M\,dx+N\,dy}$
${y'=F\left(\frac{y}{x}\right)}$	Substitute ${v=\frac{y}{x}}$
${yF(xy)\,dx+xG(xy)\,dy=0}$	Substitute ${v=xy}$
${y''+ay'+by=0}$	Solve ${m^2+am+b=0}$
${y''+ay'+by=R(x)}$	Variation of constants
${x^2y''+axy'+by=S(x)}$	Substitute ${x=e^t}$
${x^2y''+xy'+(\lambda^2x^2-n^2)y=0}$	Bessel functions $J_n,$ $Y_n$
${(1-x^2)y''}-{2xy'}+{n(n+1)y}={0}$	Legendre functions $P_n,$ $Q_n$

Separation of Variables

The equation

$$f_1(x)\,g_1(y)\,dx+f_2(x)\,g_2(y)\,dy=0$$

has coefficients that are products of a function of $x$ alone and a function of $y$ alone. Divide by ${f_2(x)\,g_1(y)}$ to collect everything with $x$ in one term and everything with $y$ in the other:

$$\frac{f_1(x)}{f_2(x)}\,dx+\frac{g_2(y)}{g_1(y)}\,dy=0.$$

An equation written as ${y'=f(x)\,g(y)}$ is of the same type: ${\frac{dy}{g(y)}=f(x)\,dx}.$ Dividing by ${g_1(y)}$ can lose solutions. Every root $y_0$ of ${g_1(y)=0}$ gives a constant solution ${y=y_0},$ so check these separately.

Example 1. Solve ${y'=xy}.$

Separate the variables: ${\frac{dy}{y}=x\,dx}.$ Integrating gives ${\ln\lvert y\rvert=\frac{x^2}{2}+c},$ so ${\lvert y\rvert=e^c\,e^{x^2/2}}.$ Put ${C=\pm e^c};$ the value ${C=0}$ gives the lost solution ${y=0}.$ The general solution is

$$y=Ce^{x^2/2}.$$

Check: ${y'=Cxe^{x^2/2}=xy}.$

Linear First-Order Equation

$$\frac{dy}{dx}+P(x)\,y=Q(x)$$

The unknown $y$ and its derivative appear only to the first power. Multiply both sides by the integrating factor ${e^{\int P\,dx}}.$ Its derivative is ${P\,e^{\int P\,dx}},$ so by the product rule the left side becomes the derivative of a product:

$$\left(y\,e^{\int P\,dx}\right)'=Q\,e^{\int P\,dx}.$$

Integrating both sides gives the solution.

Example 2. Solve ${y'+y=x}$ with the condition ${y(0)=2}.$

Here ${P=1},$ ${Q=x}$ and ${e^{\int P\,dx}=e^x}.$ Integrating by parts:

${y\,e^x=\int x\,e^x\,dx+c}={(x-1)\,e^x+c},$

so ${y=x-1+ce^{-x}}.$ The condition gives ${y(0)=-1+c=2},$ hence ${c=3}$ and

$$y=x-1+3e^{-x}.$$

Bernoulli's Equation

$$\frac{dy}{dx}+P(x)\,y=Q(x)\,y^n$$

For ${n=0}$ the equation is linear. For ${n\ne1}$ divide by $y^n$ and put ${v=y^{1-n}}.$ Since ${v'=(1-n)\,y^{-n}y'},$ the equation becomes linear in $v\text{:}$

$$v'+(1-n)P\,v=(1-n)Q.$$

Solving it as in the previous section gives the formula.

For ${n\gt0}$ the function ${y=0}$ is also a solution.

Example 3. Solve ${y'+y=y^2}.$

Here ${n=2},$ ${P=Q=1},$ ${1-n=-1}$ and ${v=y^{-1}=\frac{1}{y}}.$ The formula gives

${\frac{1}{y}\,e^{-x}=-\int e^{-x}\,dx+c}={e^{-x}+c},$

so ${\frac{1}{y}=1+ce^x}$ and

$$y=\frac{1}{1+ce^x}.$$

Exact Equation

$$M(x,y)\,dx+N(x,y)\,dy=0$$

The equation is called exact when ${\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}.$ This condition means that the left side is the total differential ${dF=\frac{\partial F}{\partial x}dx+\frac{\partial F}{\partial y}dy}$ of some function ${F(x,y)},$ with ${\frac{\partial F}{\partial x}=M}$ and ${\frac{\partial F}{\partial y}=N}.$ The equation says ${dF=0},$ so the solutions are the curves ${F(x,y)=c}.$ To find $F,$ integrate $M$ with respect to $x,$ then add the part of $N$ that this integral does not produce.

Example 4. Solve ${(2xy+3)\,dx+(x^2+4y)\,dy=0}.$

Test: ${\frac{\partial M}{\partial y}=2x}$ and ${\frac{\partial N}{\partial x}=2x},$ so the equation is exact. Then ${\int M\,\partial x=x^2y+3x}.$ Its derivative with respect to $y$ is $x^2,$ so ${N-x^2=4y}$ and ${\int4y\,dy=2y^2}.$ The solution is

$$x^2y+3x+2y^2=c.$$

Homogeneous Equation

$$\frac{dy}{dx}=F\left(\frac{y}{x}\right)$$

The right side depends only on the ratio ${\frac{y}{x}}.$ This happens, for example, when the right side is a fraction whose numerator and denominator are homogeneous polynomials of the same degree: divide both by the highest power of $x.$ Substitute ${y=vx}.$ Then ${y'=v+xv'},$ the equation becomes ${v+x\frac{dv}{dx}=F(v)},$ and the variables separate: ${\frac{dv}{F(v)-v}=\frac{dx}{x}}.$

Example 5. Solve ${y'=\frac{x+y}{x}}.$

Write the right side as ${1+\frac{y}{x}}.$ Then ${F(v)=1+v}$ and ${F(v)-v=1},$ so ${\ln\lvert x\rvert=\int dv+c=v+c}.$ Hence ${\frac{y}{x}=\ln\lvert x\rvert-c}.$ With ${C=-c}\text{:}$

$$y=x\left(\ln\lvert x\rvert+C\right).$$

Equations of the Form yF(xy)dx + xG(xy)dy = 0

$$y\,F(xy)\,dx+x\,G(xy)\,dy=0$$

Substitute ${v=xy}.$ Then ${y=\frac{v}{x}}$ and ${dy=\frac{x\,dv-v\,dx}{x^2}}.$ After simplification the equation becomes ${G(v)\,dv=\frac{v}{x}\bigl(G(v)-F(v)\bigr)\,dx},$ and the variables separate.

The special case is easy to see directly: when ${G=F},$ the equation reads ${F(xy)\,(y\,dx+x\,dy)}={F(xy)\,d(xy)}={0}.$

Example 6. Solve ${y\,dx+x(1+xy)\,dy=0}.$

Here ${F(v)=1}$ and ${G(v)=1+v},$ so ${G(v)-F(v)=v}\text{:}$

${\ln\lvert x\rvert=\int\frac{1+v}{v^2}\,dv+c}={-\frac{1}{v}+\ln\lvert v\rvert+c}.$

Put back ${v=xy}$ and use ${\ln\lvert xy\rvert=\ln\lvert x\rvert+\ln\lvert y\rvert}.$ The terms ${\ln\lvert x\rvert}$ cancel, and with ${C=-c}$ we get

$$\ln\lvert y\rvert=\frac{1}{xy}+C.$$

Linear Homogeneous Second-Order Equation

$$\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=0$$

Here $a$ and $b$ are real constants. Look for a solution of the form ${y=e^{mx}}.$ Then ${y'=me^{mx}}$ and ${y''=m^2e^{mx}}.$ Substituting and dividing by ${e^{mx}\ne0}$ gives the characteristic equation

$$m^2+am+b=0.$$

Let $m_1$ and $m_2$ be its roots. There are three cases, depending on the sign of the discriminant ${a^2-4b}.$

In Case 2 the second solution $xe^{m_1x}$ is needed because $e^{m_1x}$ alone gives only one constant. In Case 3 the complex exponentials ${e^{(p\pm qi)x}}$ are combined, by Euler's formula, into the real functions ${e^{px}\cos qx}$ and ${e^{px}\sin qx}.$

Example 7. Solve the three equations.

a) ${y''-3y'+2y=0}.$ The characteristic equation ${m^2-3m+2=0}$ has the roots $1$ and $2,$ so ${y=c_1e^x+c_2e^{2x}}.$

b) ${y''+2y'+y=0}.$ The characteristic equation ${(m+1)^2=0}$ has the double root $-1,$ so ${y=(c_1+c_2x)\,e^{-x}}.$

c) ${y''+2y'+5y=0}.$ The roots of ${m^2+2m+5=0}$ are ${-1\pm2i},$ so ${p=-1},$ ${q=2}$ and ${y=e^{-x}\left(c_1\cos2x+c_2\sin2x\right)}.$

Linear Nonhomogeneous Second-Order Equation

$$\frac{d^2y}{dx^2}+a\frac{dy}{dx}+by=R(x)$$

Here $a$ and $b$ are real constants. The general solution is the general solution of the homogeneous equation (the one with ${R=0}$) plus any one particular solution of the full equation. The method of variation of constants gives a particular solution for every right side $R(x).$ The three cases are the same as in the previous section.

When $R(x)$ is a polynomial, an exponential function, a sine or a cosine, a particular solution can often be found faster by guessing one of the same shape and finding its coefficients (the method of undetermined coefficients). The formulas above work for any $R(x),$ as Example 9 shows.

Example 8. Solve ${y''-3y'+2y=e^{3x}}.$

From Example 7a, ${m_1=1}$ and ${m_2=2}.$ The two integrals in Case 1 give

${\frac{e^x}{1-2}\int e^{-x}e^{3x}\,dx}={-e^x\cdot\frac{e^{2x}}{2}}={-\frac{e^{3x}}{2}},$

${\frac{e^{2x}}{2-1}\int e^{-2x}e^{3x}\,dx}={e^{2x}\cdot e^x}={e^{3x}}.$

Their sum is ${\frac{e^{3x}}{2}},$ so

$$y=c_1e^x+c_2e^{2x}+\frac{e^{3x}}{2}.$$

Check the particular solution: ${\left(9-9+2\right)\frac{e^{3x}}{2}=e^{3x}}.$

Example 9. Solve ${y''+y=\frac{1}{\cos x}}.$

The characteristic equation ${m^2+1=0}$ has the roots ${\pm i},$ so ${p=0}$ and ${q=1}.$ Case 3 gives

${\sin x\int\frac{\cos x}{\cos x}\,dx=x\sin x},$

${-\cos x\int\frac{\sin x}{\cos x}\,dx=\cos x\ln\lvert\cos x\rvert},$

so

${y=c_1\cos x+c_2\sin x}+{x\sin x+\cos x\ln\lvert\cos x\rvert}.$

Here guessing would not help: ${\frac{1}{\cos x}}$ is not of the shapes listed above.

Euler or Cauchy Equation

$$x^2\frac{d^2y}{dx^2}+ax\frac{dy}{dx}+by=S(x)$$

Each derivative is multiplied by the matching power of $x.$ For ${x\gt0}$ put ${x=e^t},$ that is, ${t=\ln x}.$ Then ${x\frac{dy}{dx}=\frac{dy}{dt}}$ and ${x^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}-\frac{dy}{dt}}.$

For a homogeneous equation ${(S=0)}$ you can also look for ${y=x^m}$ directly. This gives ${m(m-1)+am+b=0},$ the same characteristic equation.

Example 10. Solve ${x^2y''-2xy'+2y=x^3}$ for ${x\gt0}.$

Here ${a=-2}$ and ${b=2}.$ With ${x=e^t}$ the equation becomes ${\frac{d^2y}{dt^2}-3\frac{dy}{dt}+2y=e^{3t}},$ which is Example 8 with $t$ in place of $x\text{:}$ ${y=c_1e^t+c_2e^{2t}+\frac{e^{3t}}{2}}.$ Since ${e^t=x},$

$$y=c_1x+c_2x^2+\frac{x^3}{2}.$$

Bessel's Equation

$$x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+(\lambda^2x^2-n^2)\,y=0$$

Here $J_n$ and $Y_n$ are the Bessel functions of the first and second kind of order $n.$ They are not elementary functions. The function $J_n$ is given by the series

${J_n(x)=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!\,\Gamma(k+n+1)}\left(\frac{x}{2}\right)^{2k+n}},$

where $\Gamma$ is the gamma function (${\Gamma(m+1)=m!}$ for whole numbers $m$). The function $Y_n$ is a second, independent solution that is unbounded near ${x=0}.$ When $n$ is not an integer, ${J_{-n}(\lambda x)}$ can be used instead of ${Y_n(\lambda x)}.$ For some orders the Bessel functions are elementary, for example ${J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\,\sin x}.$

Bessel's equation appears when the wave equation or the heat equation is solved in cylindrical coordinates, for example for the vibrations of a round drum.

Transformed Bessel's Equation

${x^2\frac{d^2y}{dx^2}+(2p+1)x\frac{dy}{dx}}+{(\alpha^2x^{2r}+\beta^2)\,y}={0}$

With ${p=0},$ ${r=1},$ ${\alpha=\lambda}$ and ${\beta^2=-n^2}$ this is Bessel's equation: then ${q=n}$ and the formula gives ${y=c_1J_n(\lambda x)+c_2Y_n(\lambda x)}.$

Example 11. Solve ${x^2y''+3xy'+(4x^4+1)\,y=0}.$

Here ${2p+1=3},$ ${\alpha^2=4},$ ${2r=4}$ and ${\beta^2=1},$ so ${p=1},$ ${\alpha=2},$ ${r=2}$ and ${q=\sqrt{1-1}=0}.$ Since ${\frac{\alpha}{r}=1},$

$$y=\frac{1}{x}\left[c_1J_0(x^2)+c_2Y_0(x^2)\right].$$

Legendre's Equation

${(1-x^2)\frac{d^2y}{dx^2}}-{2x\frac{dy}{dx}}+{n(n+1)\,y}={0}$

$P_n$ and $Q_n$ are the Legendre functions of the first and second kind. For ${n=0,1,2,\ldots}$ the function $P_n$ is a polynomial of degree $n,$ the Legendre polynomial:

${P_0(x)=1},\quad\allowbreak{P_1(x)=x},\quad\allowbreak{P_2(x)=\frac{3x^2-1}{2}},\quad\allowbreak{P_3(x)=\frac{5x^3-3x}{2}}.$

The functions $Q_n$ have logarithmic singularities at ${x=\pm1},$ for example

${Q_0(x)=\frac{1}{2}\ln\frac{1+x}{1-x}},\qquad\allowbreak{Q_1(x)=\frac{x}{2}\ln\frac{1+x}{1-x}-1}.$

Check for ${n=2}\text{:}$ with ${y=P_2(x)}$ we have ${y'=3x},$ ${y''=3},$ and ${3(1-x^2)-6x^2+3(3x^2-1)=0}.$ Legendre's equation appears when Laplace's equation is solved in spherical coordinates, for example for the gravitational or electric field of a body.

Exercises

1. Solve ${y'=y\cos x}.$

Separate the variables: ${\frac{dy}{y}=\cos x\,dx},$ so ${\ln\lvert y\rvert=\sin x+c}.$ Answer: ${y=Ce^{\sin x}}.$

2. Solve ${y'=-2y}$ with ${y(0)=3}.$

As in Exercise 1, ${y=Ce^{-2x}}.$ From ${y(0)=C=3}\text{:}$ ${y=3e^{-2x}}.$

3. Solve ${y'+2y=4x}.$

A linear equation with ${P=2},$ ${Q=4x}$ and ${e^{\int P\,dx}=e^{2x}}\text{:}$ ${y\,e^{2x}=\int4x\,e^{2x}\,dx+c}={(2x-1)\,e^{2x}+c}.$ Answer: ${y=2x-1+ce^{-2x}}.$

4. Solve ${xy'+y=x^2}$ for ${x\gt0}.$

Divide by $x\text{:}$ ${y'+\frac{y}{x}=x}.$ Then ${e^{\int dx/x}=x}$ and ${xy=\int x^2\,dx+c=\frac{x^3}{3}+c}.$ Answer: ${y=\frac{x^2}{3}+\frac{c}{x}}.$ The left side ${xy'+y}$ is already the derivative of $xy,$ which gives the same result at once.

5. Solve ${y'-y=e^x\,y^2}.$

Bernoulli's equation with ${n=2},$ ${P=-1},$ ${Q=e^x}$ and ${v=\frac{1}{y}}.$ Since ${(1-n)\int P\,dx=x},$ the formula gives ${\frac{1}{y}\,e^x=-\int e^{2x}\,dx+c}={-\frac{e^{2x}}{2}+c}.$ Answer: ${y=\frac{1}{ce^{-x}-\frac{e^x}{2}}},$ and also ${y=0}.$

6. Solve ${(3x^2+y)\,dx+(x+2y)\,dy=0}.$

${\frac{\partial M}{\partial y}=1=\frac{\partial N}{\partial x}},$ so the equation is exact. ${\int M\,\partial x=x^3+xy};$ its derivative with respect to $y$ is $x,$ and ${\int(x+2y-x)\,dy=y^2}.$ Answer: ${x^3+xy+y^2=c}.$

7. Solve ${y'=\frac{y}{x}+\frac{x}{y}}.$

A homogeneous equation with ${F(v)=v+\frac{1}{v}},$ so ${F(v)-v=\frac{1}{v}}$ and ${\ln\lvert x\rvert=\int v\,dv+c=\frac{v^2}{2}+c}.$ With ${v=\frac{y}{x}}$ and ${C=-c}\text{:}$ ${y^2=2x^2\left(\ln\lvert x\rvert+C\right)}.$

8. Solve ${y''+4y=0}.$

${m^2+4=0},$ ${m=\pm2i},$ so ${p=0},$ ${q=2}.$ Answer: ${y=c_1\cos2x+c_2\sin2x}.$

9. Solve ${y''-4y'+4y=0}.$

${m^2-4m+4=(m-2)^2=0},$ a double root ${m=2}.$ Answer: ${y=(c_1+c_2x)\,e^{2x}}.$

10. Solve ${y''-y=e^{2x}}.$

${m^2-1=0},$ ${m_1=1},$ ${m_2=-1}.$ Case 1: ${\frac{e^x}{2}\int e^{x}\,dx=\frac{e^{2x}}{2}}$ and ${\frac{e^{-x}}{-2}\int e^{3x}\,dx=-\frac{e^{2x}}{6}}.$ Answer: ${y=c_1e^x+c_2e^{-x}+\frac{e^{2x}}{3}}.$

11. Solve ${y''-2y'+y=\frac{e^x}{x}}.$

A double root ${m=1}.$ Case 2: ${xe^x\int e^{-x}\frac{e^x}{x}\,dx=xe^x\ln\lvert x\rvert}$ and ${-e^x\int xe^{-x}\frac{e^x}{x}\,dx=-xe^x}.$ The term $-xe^x$ is already contained in $c_2xe^x.$ Answer: ${y=(c_1+c_2x)\,e^x+xe^x\ln\lvert x\rvert}.$

12. Solve ${x^2y''+xy'-y=0}$ for ${x\gt0}.$

An Euler equation with ${a=1},$ ${b=-1}.$ With ${x=e^t}\text{:}$ ${\frac{d^2y}{dt^2}-y=0},$ so ${y=c_1e^t+c_2e^{-t}}.$ Answer: ${y=c_1x+\frac{c_2}{x}}.$