# Thread: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

1. ## (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

what's wrong with my working? I couldn't get the ans , btw , i have written down the answer given inside the picture.

P/ s: the correct question is (1-x^2)(dy/dx) -xy = 1/(1-x^2)

2. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

\displaystyle \begin{aligned}(1-x^2){\mathrm d y \over \mathrm d x} - xy &= {1 \over 1-x^2} \\ (1-x^2){\mathrm d y \over \mathrm d x} - xy - {1 \over 1-x^2} &= 0 &\text{exact?} \\[8pt] F_y(x,y) = N(x,y) &= 1-x^2 \implies N_x=-2x \\ F_x(x,y) = M(x,y) &= -xy-{1 \over 1-x^2} \implies M_y = -x &\text{no}\\ {M_y - N_x \over N} &= {x \over 1-x^2} \implies u(x) = {1 \over \sqrt{1-x^2}} &\text{integrating factor}\\[12pt] \sqrt{1-x^2} {\mathrm d y \over \mathrm d x} -{xy \over \sqrt{1-x^2}} - (1-x^2)^{-\frac32}&= 0 \\[8pt] F_y = \sqrt{1-x^2} \implies F(x,y) &= y\sqrt{1-x^2} + f(x) \\ F_x = {-xy \over \sqrt{1-x^2}} + f'(x) \implies f'(x) &= (1-x^2)^{-\frac32} \\ \implies f(x) &= \int {\mathrm d x \over (1-x^2)^\frac32} = \int {\cos u \, \mathrm d u \over \cos^3 u} & (x = \sin u) \\ &= \int \sec^2 u \, \mathrm d u = \tan u + c \\ &= {x \over \sqrt{1-x^2}} \\ F(x,y) &= y\sqrt{1-x^2} + {x \over \sqrt{1-x^2}} = c \\[12pt] y &= {c \over \sqrt{1-x^2}} - {x \over 1-x^2}\end{aligned}

You should check the working.
I think your attempt went wrong in the integral of $\displaystyle (1-x^2)^{-\frac32}$ where you should have done the substitution.

3. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

Originally Posted by Archie
\displaystyle \begin{aligned}(1-x^2){\mathrm d y \over \mathrm d x} - xy &= {1 \over 1-x^2} \\ (1-x^2){\mathrm d y \over \mathrm d x} - xy - {1 \over 1-x^2} &= 0 &\text{exact?} \\[8pt] F_y(x,y) = N(x,y) &= 1-x^2 \implies N_x=-2x \\ F_x(x,y) = M(x,y) &= -xy-{1 \over 1-x^2} \implies M_y = -x &\text{no}\\ {M_y - N_x \over N} &= {x \over 1-x^2} \implies u(x) = {1 \over \sqrt{1-x^2}} &\text{integrating factor}\\[12pt] \sqrt{1-x^2} {\mathrm d y \over \mathrm d x} -{xy \over \sqrt{1-x^2}} - (1-x^2)^{-\frac32}&= 0 \\[8pt] F_y = \sqrt{1-x^2} \implies F(x,y) &= y\sqrt{1-x^2} + f(x) \\ F_x = {-xy \over \sqrt{1-x^2}} + f'(x) \implies f'(x) &= (1-x^2)^{-\frac32} \\ \implies f(x) &= \int {\mathrm d x \over (1-x^2)^\frac32} = \int {\cos u \, \mathrm d u \over \cos^3 u} & (x = \sin u) \\ &= \int \sec^2 u \, \mathrm d u = \tan u + c \\ &= {x \over \sqrt{1-x^2}} \\ F(x,y) &= y\sqrt{1-x^2} + {x \over \sqrt{1-x^2}} = c \\[12pt] y &= {c \over \sqrt{1-x^2}} - {x \over 1-x^2}\end{aligned}

You should check the working.
I think your attempt went wrong in the integral of $\displaystyle (1-x^2)^{-\frac32}$ where you should have done the substitution.
is it wrong to do the integration$\displaystyle (1-x^2)^{-\frac32}$ straight away without using the substituition ?

4. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

It's fine if you get the right answer.

5. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

Originally Posted by Archie
It's fine if you get the right answer.
can you show how to integrate (1-x^2) ^-1.5 without substituition ? i have tried many times , but i get sqrt rt (1-x^2) / -x , here 's my working

6. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

Well I would use a substitution to get the right answer. If you differentiate the answer, you will get an idea as to why it might be so difficult. Probably the key to direct integration would be to multiply by $\displaystyle {x \over x}$ and then integrate by parts.

But the point of having different techniques is that sometimes one or more of them is easier than the rest. The idea is to use the easiest technique for the context, not to stick rigidly to one technique and try to use it everywhere.

7. ## Re: (1-x^2)_(dy/dx) -xy = 1/(1-x^2)

Originally Posted by xl5899
what's wrong with my working? I couldn't get the ans , btw , i have written down the answer given inside the picture.

P/ s: the correct question is (1-x^2)(dy/dx) -xy = 1/(1-x^2)
Since you have attempted using an integrating factor, as it's first order linear, I will follow that method...

\displaystyle \begin{align*} \left( 1 - x^2 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} - x\,y &= \frac{1}{1 - x^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{x}{1 - x^2}\,y &= \frac{1}{\left( 1 - x^2 \right) ^2 } \end{align*}

Now the integrating factor is \displaystyle \begin{align*} \mathrm{e}^{\int{ -\frac{x}{1 - x^2} \,\mathrm{d}x }} = \mathrm{e}^{\int{\frac{1}{2}\,\left( \frac{-2x}{1 - x^2} \right)\,\mathrm{d}x}} = \mathrm{e}^{ \frac{1}{2}\ln{ \left( 1 - x^2 \right) } } = \mathrm{e}^{ \ln{\left[ \left( 1 - x^2 \right) ^{\frac{1}{2}} \right] } } = \left( 1 - x^2 \right) ^{\frac{1}{2}} \end{align*} so multiplying both sides by the integrating factor gives

\displaystyle \begin{align*} \left( 1 - x^2 \right) ^{\frac{1}{2}} \,\frac{\mathrm{d}y}{\mathrm{d}x} - \frac{x}{\left( 1 - x^2 \right) ^{\frac{1}{2}}} \, y &= \frac{1}{ \left( 1 - x^2 \right) ^{\frac{3}{2}}} \\ \frac{\mathrm{d}}{\mathrm{d}x} \, \left[ \left( 1 - x^2 \right) ^{\frac{1}{2}}\,y \right] &= \frac{1}{ \left( 1 - x^2 \right) ^{\frac{3}{2}}} \\ \left( 1 - x^2 \right) ^{\frac{1}{2}}\,y &= \int{ \frac{1}{ \left( 1 - x^2 \right) ^{\frac{3}{2}}} \,\mathrm{d}x } \\ \left( 1 - x^2 \right) ^{\frac{1}{2}}\,y &= \int{ \frac{1}{\left[ 1 - \sin^2{(t)} \right] ^{\frac{3}{2}}}\,\cos{(t)}\,\mathrm{d}t } \textrm{ after substituting } x = \sin{(t)} \implies \mathrm{d}x = \cos{(t)}\,\mathrm{d}t \\ \left( 1 - x^2 \right) ^{\frac{1}{2}} \,y &= \int{ \frac{ \cos{(t)} }{ \left[ \cos^2{(t)} \right] ^{\frac{3}{2}} } \,\mathrm{d}t } \\ \left( 1 - x^2 \right) ^{ \frac{1}{2} } \,y &= \int{ \frac{1}{\cos^2{(t)}} \,\mathrm{d}t }\\ \left( 1 - x^2 \right) ^{\frac{1}{2}}\,y &= \tan{(t)} + C \\ \left( 1 - x^2 \right) ^{\frac{1}{2}}\,y &= \frac{\sin{(t)}}{ \left [ 1 - \sin^2{(t)} \right] ^{\frac{1}{2}} } + C \\ \left( 1 - x^2 \right) ^{\frac{1}{2}} \,y &= \frac{x}{ \left( 1 - x^2 \right) ^{\frac{1}{2}} } + C \\ y &= \frac{x}{ 1 - x^2 } + \frac{C}{ \left( 1 - x^2 \right) ^{\frac{1}{2}} } \end{align*}