What is the solution for (36-x^2)(dy)/(dx)=1 ?

The solution for (36-x^2)(dy)/(dx)=1 is y= 1/12 (ln(x/6+1)-ln(x/6-1))+c_{1}

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(36-x^2)(dy)/(dx)=1

(36 - x^{2}) \frac{d y}{d x} = 1

y = \frac{1}{12} (ln (\frac{x}{6} + 1) - ln (\frac{x}{6} - 1)) + c_{1}

Solution steps

(36 - x^{2}) \frac{d y}{d x} = 1

Substitute

\frac{d y}{d x}

with

y^{'} (x)

(36 - x^{2}) y^{^{'}} (x) = 1

Isolate

y^{'} (x) : y^{'} (x) = \frac{1}{36 - x ^{2}}

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

then

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

y = \int \frac{1}{36 - x ^{2}} d x

\int \frac{1}{36 - x ^{2}} d x = \frac{1}{12} (ln (\frac{x}{6} + 1) - ln (\frac{x}{6} - 1)) + c_{1}

y = \frac{1}{12} (ln (\frac{x}{6} + 1) - ln (\frac{x}{6} - 1)) + c_{1}

\int_{0}^{\frac{π}{3}} (tan (x))^{2} d x

\int 9 ln (x^{2} - 1) d x

y^{^{'}} + \frac{y}{x} = \frac{y ^{2}}{x}

\frac{d}{d x} (x^{4} - 5 x^{2} + 4)

y = \frac{4}{( 4 - x ^{4} ) ^{\frac{3}{2}}}

What is the solution for (36-x^2)(dy)/(dx)=1 ?
The solution for (36-x^2)(dy)/(dx)=1 is y= 1/12 (ln(x/6+1)-ln(x/6-1))+c_{1}