# Differential Equations of Form dy/dx = f(x) or f(y)

Here you will learn how to find the general solution of differential equations of form dy/dx = f(x) or f(y) with examples.

Let’s begin –

## Differential Equations of Form dy/dx = f(x) or f(y)

#### (1) Differential Equations of Form dy/dx = f(x)

To solve this type of differential equations we integrate both sides to obtain the general solution as discussed below.

We have.

$$dy\over dx$$ = f(x) $$\iff$$ dy = f(x)dx

Integrating both sides, we obtain

$$\int$$ dy = $$\int$$ f(x) dx + C or,

y = $$\int$$ f(x) dx + C,

which gives general solution of the differential equation.

Example : Solve the given differential equation : $$dy\over dx$$ = $$x\over x^2 + 1$$

Solution : We have,

$$dy\over dx$$ = $$x\over x^2 + 1$$

$$\implies$$ dy = $$x\over x^2 + 1$$dx

Integrating both sides, we get

$$\int$$ dy = $$\int$$ $$x\over x^2 + 1$$dx

$$\implies$$ dy = $$1\over 2$$ $$2x\over x^2 + 1$$dx

$$\implies$$ y = $$1\over 2$$ $$log|x^2 + 1|$$ + C

Clearly, y = $$1\over 2$$ $$log|x^2 + 1|$$ + C is defined for all x $$\in$$ R.

Hence, y = $$1\over 2$$ $$log|x^2 + 1|$$ + C, x $$\in$$ R is the solution of the given differential equation.

#### (2) Differential Equations of Form dy/dx = f(y)

To solve this type of differential equations we integrate both sides to obtain the general solution as discussed below.

We have.

$$dy\over dx$$ = f(y)

$$\implies$$ $$dx\over dy$$ = $$1\over f(y)$$, provided that f(y) $$\ne$$ 0

$$\implies$$ dx = $$1\over f(y)$$ dy

Integrating both sides, we obtain

$$\int$$ dx = $$\int$$ $$1\over f(y)$$ dy + C or,

x = $$\int$$ $$1\over f(y)$$ dy + C,

which gives general solution of the differential equation.

Example : Solve the given differential equation :  $$dy\over dx$$ = $$1\over y^2 + sin y$$

Solution : We have,

$$dy\over dx$$ = $$1\over y^2 + sin y$$

$$\implies$$ $$dx\over dy$$ = $$y^2 + sin y$$

$$\implies$$ dx = $$y^2 + sin y$$dx

Integrating both sides, we get

$$\int$$ dx = $$\int$$ $$y^2 + sin y$$dx

$$\implies$$ x = $$y^3\over 3$$ – cosy + C

Hence, x = $$y^3\over 3$$ – cosy + C is the solution of the given differential equation.