What is the solution for 2x^3\H(x)(x)^2dx+x^4\H(x)d\H(x)=0 ?

The solution for 2x^3\H(x)(x)^2dx+x^4\H(x)d\H(x)=0 is f(x)=-x^2+c_{1}

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2x^3\H(x)(x)^2dx+x^4\H(x)d\H(x)=0

2 x^{3} H (x) (x)^{2} d x + x^{4} H (x) d H (x) = 0

f (x) = - x^{2} + c_{1}

Solution steps

2 x^{3} H (x) x^{2} d x + x^{4} H (x) d H (x) = 0

Let

H (x)

be the dependent variable. Divide by

d x

2 x^{3} H (x) x^{2} + x^{4} H (x) \frac{d H ( x )}{d x} = 0

Substitute

\frac{d H ( x )}{d x}

with

f^{'} (x)

2 x^{3} f (x) x^{2} + x^{4} f (x) f^{^{'}} (x) = 0

Isolate

f^{'} (x) : f^{'} (x) = - 2 x

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

f (x) = \int - 2 x d x

\int - 2 x d x = - x^{2} + c_{1}

f (x) = - x^{2} + c_{1}

(1 + x^{2}) \cdot y^{^{'}} = x (1 - y)

\frac{d y}{d x} - 2 x y = 4 x, y (0) = 3

\frac{d y}{d x} = y \cdot sin (x)

y (x^{3} d y + y^{3} d x) = 0

\frac{d V}{d t} = 4 kπ (\frac{3 V}{4 π})^{\frac{2}{3}}

What is the solution for 2x^3\H(x)(x)^2dx+x^4\H(x)d\H(x)=0 ?
The solution for 2x^3\H(x)(x)^2dx+x^4\H(x)d\H(x)=0 is f(x)=-x^2+c_{1}