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Solve Differential Equation

Solve a differential equation analytically by using the dsolve function, with or without initial conditions. To solve a system of differential equations, see Solve a System of Differential Equations.

First-Order Linear ODE

Solve this differential equation.

dydt=ty.

First, represent y by using syms to create the symbolic function y(t).

syms y(t)

Define the equation using == and represent differentiation using the diff function.

ode = diff(y,t) == t*y
ode(t) =
diff(y(t), t) == t*y(t)

Solve the equation using dsolve.

ySol(t) = dsolve(ode)
ySol(t) =
C1*exp(t^2/2)

Solve Differential Equation with Condition

In the previous solution, the constant C1 appears because no condition was specified. Solve the equation with the initial condition y(0) == 2. The dsolve function finds a value of C1 that satisfies the condition.

cond = y(0) == 2;
ySol(t) = dsolve(ode,cond)
ySol(t) =
2*exp(t^2/2)

If dsolve cannot solve your equation, then try solving the equation numerically. See Solve a Second-Order Differential Equation Numerically.

Nonlinear Differential Equation with Initial Condition

Solve this nonlinear differential equation with an initial condition. The equation has multiple solutions.

(dydt+y)2=1,y(0)=0.

syms y(t)
ode = (diff(y,t)+y)^2 == 1;
cond = y(0) == 0;
ySol(t) = dsolve(ode,cond)
ySol(t) =
 exp(-t) - 1
 1 - exp(-t)

Second-Order ODE with Initial Conditions

Solve this second-order differential equation with two initial conditions.

d2ydx2=cos(2x)y,y(0)=1,y'(0)=0.

Define the equation and conditions. The second initial condition involves the first derivative of y. Represent the derivative by creating the symbolic function Dy = diff(y) and then define the condition using Dy(0)==0.

syms y(x)
Dy = diff(y);

ode = diff(y,x,2) == cos(2*x)-y;
cond1 = y(0) == 1;
cond2 = Dy(0) == 0;

Solve ode for y. Simplify the solution using the simplify function.

conds = [cond1 cond2];
ySol(x) = dsolve(ode,conds);
ySol = simplify(ySol)
ySol(x) =
1 - (8*sin(x/2)^4)/3

Third-Order ODE with Initial Conditions

Solve this third-order differential equation with three initial conditions.

d3udx3=u,u(0)=1,u(0)=1,u(0)=π.

Because the initial conditions contain the first- and second-order derivatives, create two symbolic functions, Du = diff(u,x) and D2u = diff(u,x,2), to specify the initial conditions.

syms u(x)
Du = diff(u,x);
D2u = diff(u,x,2);

Create the equation and initial conditions, and solve it.

ode = diff(u,x,3) == u;
cond1 = u(0) == 1;
cond2 = Du(0) == -1;
cond3 = D2u(0) == pi;
conds = [cond1 cond2 cond3];

uSol(x) = dsolve(ode,conds)
uSol(x) =
 
(pi*exp(x))/3 - exp(-x/2)*cos((3^(1/2)*x)/2)*(pi/3 - 1) -...
(3^(1/2)*exp(-x/2)*sin((3^(1/2)*x)/2)*(pi + 1))/3

More ODE Examples

This table shows examples of differential equations and their Symbolic Math Toolbox™ syntax.

Differential Equation

MATLAB® Commands

dydt+4y(t)=et,y(0)=1.

syms y(t)
ode = diff(y)+4*y == exp(-t);
cond = y(0) == 1;
ySol(t) = dsolve(ode,cond)
ySol(t) =
exp(-t)/3 + (2*exp(-4*t))/3

2x2d2ydx2+3xdydxy=0.

syms y(x)
ode = 2*x^2*diff(y,x,2)+3*x*diff(y,x)-y == 0;
ySol(x) = dsolve(ode)
ySol(x) =
C1/(3*x) + C2*x^(1/2)

The Airy equation.

d2ydx2=xy(x).

syms y(x)
ode = diff(y,x,2) == x*y;
ySol(x) = dsolve(ode)
ySol(x) =
C1*airy(0,x) + C2*airy(2,x)

Puiseux series solution.

(x2+1)d2ydx22xdydx+y=0.

syms y(x) a
ode = (x^2+1)*diff(y,x,2)-2*x*diff(y,x)+y == 0;
Dy = diff(y,x);
cond = [Dy(0) == a; y(0) == 5];
ySol(x) = dsolve(ode,cond,'ExpansionPoint',0)
ySol(x) =
- (a*x^5)/120 - (5*x^4)/24 + (a*x^3)/6 - (5*x^2)/2 + a*x + 5

See Also

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