Hi!
I have these set of problems that i am finding hard to solve. Can you please provide me the solutions of them?
Thank you.
These are all pretty much separable or separable with a simple substitution. Is it the integrals that a blocking you?
For #2 you have (1+ x^2)dy= (7sqrt{1- y^2})x. That obviously separates as [math]\frac{dy}{\sqrt{1- y^2}= \frac{dx}{1+ x^2}[/tex]. Those are both "inverse trig function" integrals that you should know from Calculus. $\displaystyle \int \frac{dy}{\sqrt{1- x^2}}= arcsin(y)+ C_1$. $\displaystyle \int \frac{dx}{1+ x^2}= arctan(x)+ C_2$. So the solution to the differential equation is $\displaystyle arcsin(y)+ C_1= arctan(x)+ C_2$. You can combine the two constants of integration to get $\displaystyle arcsin(y)= arctan(x)+ C$ and, if you wish, solve for y: $\displaystyle y= sin(arctan(x)+ C)$.
#3 is $\displaystyle xdy- ydx= \sqrt{1+ x^2}dy+ \sqrt{1+ y^2}dx$. Again, this is separable as $\displaystyle \frac{dy}{y+ \sqrt{1+ y^2}}= \frac{dx}{x- \sqrt{1+ x^2}}$. Can you integrate that?
#4 is y'= (y- x)^2+ 1. A pretty obvious substitution is to let u= y- x. Then y= u+ x so y'= u'+ 1. The equation becomes $\displaystyle u'+ 1=u^2+ 1$ or $\displaystyle u'= u^2$ so $\displaystyle u^{-2}du= dx$. Integrate!
#5 is $\displaystyle 2x\sqrt{4- x^2}+ yy'= 0$ so $\displaystyle ydy= -2x\sqrt{4- x^2}dx$. To integrate on the right hand side, let $\displaystyle u= 4- x^2$ so that $\displaystyle du= -2x dx$.
#6 is $\displaystyle x' sin(t)= x ln(x)$ which immediately separates as $\displaystyle \frac{dx}{x ln(x)}= \frac{1}{sin(t)}dt=csc(t)dt$. On the left let u=ln(x). The right is a standard trig integral.
#7 is just $\displaystyle \frac{t^3+ 1}{t}dt= \frac{x^2- 1}{x}dx$. That's just basic algebra! $\displaystyle \frac{t^2+ 1}{t}= t^2+ \frac{1}{t}$ and $\displaystyle \frac{x^2- 1}{x}= x- \frac{1}{x}$ and those should be easy to integrate.
#8 is $\displaystyle (1+ 2e^{x/y})dx+ 2e^{x/y}\left(1-\frac{x}{y}\right)dy= 0$. Seeing those three 'x/y' terms the substitution u= x/y should leap right out at you. Then x= uy so dx= udy+ udu. The equation becomes $\displaystyle (1+ 2e^u)(udy+ ydu)+ 2e^u(1- u)dy= (u- 2e^u)dy= -(u+ 2e^u)ydu$. $\displaystyle \frac{dy}{y}= -\frac{u+ 2e^u}{u- 2e^u}du$.
No, I am not going to go through all 29 (!) problems. As I said to begin with, these are all either separable or separable with a simple substitution- you learn by doing, not by asking some one else to do them for you!