Extraordinary Differential Equations #5

Relevant wiki: Variable changes in linear differential equations of first order: y'=f(t,y)

Instead of coupling the $x$ and $y$ terms together within the same substitution $V=x+y+\alpha$ into the ODE for some constant $\alpha$ , notice that this method does not work here because there are no viable values for $\alpha$ such that the ODE becomes separable or homogeneous.

Therefore, we try using two substitutions instead: $X=x+\alpha$ and $Y=y+\beta$ and find suitable values for real-valued constants $\alpha$ and $\beta$ such that the ODE becomes homogeneous (notice that also $y'=Y'$ ): $X-Y-\alpha+\beta-1+(X+Y-\alpha-\beta+2)y'=0 \rightarrow X-Y+(X+Y)Y'=0$ which means we need to solve $-\alpha+\beta-1=0$ and $-\alpha-\beta+2=0$ which gives $\alpha=1/2$ and $\beta=3/2$ .

Using the substitution $W=Y/X \implies Y'=W+XW'$ we get $1-W+(1+W)(W+XW')=0$ the solution to which is $\frac{1}{2} \ln{(W^2+1)}+\tan^{-1} W=c_1-\ln{X}$ Back substitution of the original terms gives $\frac{1}{2} \ln{\left(\left(\frac{y+1.5}{x+0.5}\right)^2+1\right)}+\tan^{-1} \left(\frac{y+3/2}{x+1/2}\right)=k_1-\ln{\left(x+\frac{1}{2}\right)}$ Therefore $A+B+C+D=2+2+3+1=8$ .

Firstly Lets replace x by x'+h and y by y'+k

Now we adjust h and k such that constant terms vanishes and we get a homogeneous differential equation

doing so we obtain h = -0.5 and k = -1.5

Now substitute y' = vx' which is a standard substitution for solving homogeneous differential equations .

on solving we get the required answer

it can be done without pen and paper and i did so, observe carefully , you would see how ! since the given form contains y+1.5 ad x+0.5 all over , it was a hint what would be C=3 , D=1 so , you need not do what bro @Prakhar Bindal has done if you watch closely and do some calculation (again in mind:P) you would get A=B=2 so, i recommend you too to do the questions in mind without making yourself work to get paper and pen , laziness rocks !