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solmaris [256]
3 years ago
6

Find the solution of the problem (1 3. (2 cos x - y sin x)dx + (cos x + sin y)dy=0.

Mathematics
1 answer:
lakkis [162]3 years ago
7 0

Answer:

2*sin(x)+y*cos(x)-cos(y)=C_1

Step-by-step explanation:

Let:

P(x,y)=2*cos(x)-y*sin(x)

Q(x,y)=cos(x)+sin(y)

This is an exact differential equation because:

\frac{\partial P(x,y)}{\partial y} =-sin(x)

\frac{\partial Q(x,y)}{\partial x}=-sin(x)

With this in mind let's define f(x,y) such that:

\frac{\partial f(x,y)}{\partial x}=P(x,y)

and

\frac{\partial f(x,y)}{\partial y}=Q(x,y)

So, the solution will be given by f(x,y)=C1, C1=arbitrary constant

Now, integrate \frac{\partial f(x,y)}{\partial x} with respect to x in order to find f(x,y)

f(x,y)=\int\  2*cos(x)-y*sin(x)\, dx =2*sin(x)+y*cos(x)+g(y)

where g(y) is an arbitrary function of y

Let's differentiate f(x,y) with respect to y in order to find g(y):

\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y} (2*sin(x)+y*cos(x)+g(y))=cos(x)+\frac{dg(y)}{dy}

Now, let's replace the previous result into \frac{\partial f(x,y)}{\partial y}=Q(x,y) :

cos(x)+\frac{dg(y)}{dy}=cos(x)+sin(y)

Solving for \frac{dg(y)}{dy}

\frac{dg(y)}{dy}=sin(y)

Integrating both sides with respect to y:

g(y)=\int\ sin(y)  \, dy =-cos(y)

Replacing this result into f(x,y)

f(x,y)=2*sin(x)+y*cos(x)-cos(y)

Finally the solution is f(x,y)=C1 :

2*sin(x)+y*cos(x)-cos(y)=C_1

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