Question

Solve the intital value problem
(cosy)dy/dx-siny=x,y(0)=0

Answer 1

Answer on Question #58664 – Math – Differential Equations

Question

Solve the initial value problem

$$(\cos y) \, dy/dx \cdot \sin y = x, \, y(0) = 0$$

Solution

1) We have an equation

$$\cos y \cdot y' - \sin y = x,$$

that is,

$$(\sin y)' - \sin y = x.$$

Denote

$$u = \sin y.$$

We get a linear equation with respect to $u$ :

$$u' - u = x.$$

2) Multiplying the equation (1) by $e^{-x}$ we have

$$e^{-x} u' - e^{-x} u = x e^{-x},$$

which is equivalent to

$$(e^{-x} u)' = x e^{-x}.$$

Integrating both parts with respect to $x$

$$e^{-x} u = \int x e^{-x} \, dx + C,$$

hence

$$e^{-x} u = -x e^{-x} - e^{-x} + C$$

and multiplying the equality by $e^x$

we get

$$u = -x - 1 + C e^x.$$

3) Substituting $u = \sin x$ we have

$$\sin y = - x - 1 + C e ^ {x}$$

and using initial conditions $x = 0, y = 0$ we get $C = 1$ .

Finally obtain

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

**Answer:** $\sin y = -x - 1 + e^{x}$ .

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