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In problems 1-8 Classify the equation as separable, linear, exact, or none of these. Notice that some equations may have more than one classification.

y2dx+(2xy+cosy)dy=0

Short Answer

Expert verified

The equation is linear and also exact.

Step by step solution

01

Find if the equation is separable

Here

y2dx+(2xy+cosy)dy=0dxdy=-(2xy+cosy)y2

This equation can be written in the form as g(x) h(y).so, this equation is not separable.

02

Determine if the equation is linear

The equation can be written as

dxdy+xy=-cosyy

Now,

P(x)=1y,Q(x)=-cosyy

This equation is a linear.

03

 Step 3: Evaluate whether the equation is exact

The condition for exact is ∂M∂y=∂N∂x.

M(x,y)=y2N(x,y)=2xy+cosy∂M∂y=2y∂N∂x=2y∂M∂y=∂N∂x

This equation is exact.

Hence, the equation is linear and also exact.

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