Differential Equation Solver

Solve first-order differential equations step by step: linear, separable, Bernoulli, homogeneous and exact (or reducible to exact), with an optional initial condition and the slope field.

Type	Linear: dy/dx + P(x)y = Q(x)
P(x)	1/x
Q(x)	x^2
Initial condition: at x =	1
y =	1

Type

P(x)

Q(x)

Initial condition: at x =

Optional: leave both empty for the general solution.

y =

dy/dx + (1/x)y = x^2

y = (x^4/4 + 0.75) / x

The integrating factor is x (taking x > 0). With y(1) = 1, C = 0.75.

Exact form. Checked numerically.

Method: Integrating factor
Worked out with: AllCalcHub integration rules
Check: Every step that can be checked was checked; the solution substituted into dy/dx + P(x)y and found equal to Q(x)

Integrating factor: x
General solution: y·(x) = x^4/4 + C

Solution curves for a few values of C, on the slope field

Initial conditionat (1, 1)Point at the chart, or Tab to it and use ← →

Initial condition: (1, 1)

About the Differential Equation Solver

A linear equation dy/dx + P(x)y = Q(x) becomes easy after multiplying by the integrating factor IF = e^(∫P dx): the left side turns into d(y·IF)/dx, so y·IF = ∫Q·IF dx + C.

A separable equation dy/dx = f(x)g(y) is solved by putting the y's on one side and the x's on the other and integrating both. An initial condition y(x₀) = y₀ fixes C.

A homogeneous equation dy/dx = f(x, y), where f depends only on y/x, becomes separable with y = vx. A Bernoulli equation dy/dx + Py = Qyⁿ becomes linear with v = y^(1−n). M dx + N dy = 0 is exact when ∂M/∂y = ∂N/∂x; then its solution is φ(x, y) = C with ∂φ/∂x = M and ∂φ/∂y = N. If it isn't exact, an integrating factor e^(∫(M_y − N_x)/N dx) or e^(∫(N_x − M_y)/M dy) may make it exact.

The formula

IF = e^(∫P dx), y × IF = ∫ Q × IF dx + C

Worked example

dy/dx + y/x = x², y(1) = 1

IF = e^(ln x) = x
xy = ∫ x³ dx = x⁴/4 + C
C = 3/4, so y = (x⁴/4 + 3/4)/x

Questions people ask

What does the slope field show?

A short dash at each point with the slope dy/dx there. Every solution curve runs along the dashes; the one drawn passes through the initial condition (traced numerically).

Why take x > 0?

e^(ln|x|) = |x|; textbooks take the region where x > 0 so the integrating factor is just x.

What's the difference between general and particular solutions?

The general solution has the constant C; an initial condition picks one curve, the particular solution.