2. The general solution of x^2y' + 4xy = 4x^3 is (a) y = 4 +...

2. The general solution of $x^2y' + 4xy = 4x^3$ is (a) $y = 4 + Cx^{-4}$ (b) $y = x^2 + Cx^4$ (c) $y = frac{2}{3}x^2 + Cx^4$ (d) $y = frac{2}{3}x^2 + Cx^{-4}$ (e) None of the above. 3. The general solution of $frac{dy}{dx} = frac{xy^2 - x}{y}$ is (a) $y^2 = Ce^{x^2} + 1$ (b) $ln|y^2 - 1| = ln x + C$ (c) $ln|y^2 - 1| = Ce^{x^2}$ (d) $y^2 = Cx^2 - 1$ (e) None of the above.

Transcript

00:01 We will now solve this differential equation which is given as x equal to 1 plus x y times of d a bdx plus x squared y squared by 2 factorial times of du bdx quantity squared plus x cube plus x cube times of d a bdx cube and so on.

00:18 So to solve this differential equation, i'm going to write the series expansion of e power x.

00:23 So this is given as 1 plus x plus x squared by 2 factorial.

00:30 Plus x cube by three factorial plus and so on so this is basically the series expansion for e power x so in this i'm going to replace x as xy times of diva by d x so when i do that i'll be getting e power xy times of dy by d x so this is equal to 1 plus xy times of diva by d x plus xy times of diva by d x plus xy times of die by d x square by two factorial plus xy times of diva by d x squared by dx by dx by two factorial so if you could absorb this right side of this e power xy with the given differential equation the right side matches the same basically both the right sides are exactly the same which means their left side should also be the same same so we can write down as x equal to e power xy times of d y by d x okay so now i'm going to take natural logarithm on both sides and when i do that i will get ln x is equal to ln e to the power of xy times of d y by d x so this is equivalent to ln can write this as when we apply the property of logarithm this will get multiplied with natural logarithm this comes in front so i can relate this as xy times of d y by d x natural logarithm of e observe that the value of natural algorithm of e is basically one so i can simply write this as times of one or just this expression so we have ln x we have this equation ln of x is equal to x y times of d, y, by dx.

02:42 Now, let's separate the variables, that is, all y terms on to one of the sides, and all x terms to other side.

02:49 So, for that, i need to multiply by dx and also divide by x.

02:55 So when i do that, i'll be getting lnx, divide by x times of dx is equal to, i'll be getting y times of d, so i rewrite this in a different form, that is, i transpose the y terms to the left side then i will get y times of d -y is equal to this i write down as 1 over x times of lnx d x now we have separated the variables we can integrate both sides so let's put the integration symbol this will be integral integral and when you integrate the left side that is basically the integration of y will get y squared by 2 and to integrate this part, we make a substitution.

03:48 That is, i can put lnx equal to t, which means if i take differentials on both sides, i'll be getting 1x times of dx is equal to t...

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