What is the (1/x+2y^2x)dx+(2yx^2-cos(y))dy=0,y(1)=pi ?

The (1/x+2y^2x)dx+(2yx^2-cos(y))dy=0,y(1)=pi is 0

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(1/x+2y^2x)dx+(2yx^2-cos(y))dy=0,y(1)=pi

(\frac{1}{x} + 2 y^{2} x) d x + (2 y x^{2} - cos (y)) d y = 0, y (1) = π

0

Solution steps

For

(\frac{1}{x} + 2 y^{2} x) d x + (2 y x^{2} - cos (y)) d y \circ y (1)

substitute

(\frac{1}{x} + 2 y^{2} x) d x + (2 y x^{2} - cos (y)) d y

with

0, y (1)

with

π

= 0 \circ π

0 \circ π = 0

= 0

n = 1 \sum \infty (sin (3))^{n}

\frac{d}{d x} (e^{x + y} - 1)

\int \frac{sin ^{4} ( x )}{cos ^{6} ( x )} d x

f (x) = 7 x, at x = 4

n = 5 \sum \infty \frac{n ^{3}}{2 ^{n}}

What is the (1/x+2y^2x)dx+(2yx^2-cos(y))dy=0,y(1)=pi ?
The (1/x+2y^2x)dx+(2yx^2-cos(y))dy=0,y(1)=pi is 0