Author Topic: TUT0402 Quiz1  (Read 504 times)

Di Qiu

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TUT0402 Quiz1
« on: September 27, 2019, 02:00:04 PM »
To find the general solution of: $$ \frac{dy}{dx}x = (1-y^2)^{\frac{1}{2}} $$
Rearrange: $$ \frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \frac{1}{x} dx $$
Take integral on both sides: $$ \int \frac{1}{(1-y^2)^{\frac{1}{2}}} dy = \int \frac{1}{x} dx $$
Left-hand side: $$ \int (1-y^2)^{\frac{1}{2}} dy = arcsin y $$
Right-hand side: $$ \int \frac{1}{x} dx = ln |x| $$
Therefore the general solution: $$ arcsin y = ln|x| + C $$
Keep only y on the left-hand side: $$ y = sin (ln|x| + C), x \neq 0 $$
 
« Last Edit: September 27, 2019, 05:50:11 PM by yuki »