Question

Solve the first order differential equation
\[(5x+2y+1)\text d x+(2x+y+1)\text d y=0\]

Answer 1

\[\frac{\text d y}{\text d x}=-\frac{5x+2y+1}{2x+y+1}\]

Answer 2

\[\frac{\text d Y}{\text d X}=-\frac{5(X+p)+2(Y+q)+1}{2(X+p)+(Y+q)+1}\]\[\frac{\text d Y}{\text d X}=-\frac{5X+2Y+(5p+2q+1)}{2X+Y+(2p+q+1)}\]
\[2p+q+1=0\qquad5p+2q+1=0\]\[q=-2p-1\qquad5p+2(-2p-1)+1=0\]\[q=-2(1)-1\qquad p-1=0\]\[q=-3\qquad p=1\]
\[\frac{\text d Y}{\text d X}=-\frac{5X+2Y}{2X+Y}\]

Answer 3

\[\frac{\text d Y}{\text d X}=-\frac{5+2\frac{Y}{X}}{2+\frac{Y}{X}}\]
\[\frac{Y}{X}=V\qquad Y=VX\]\[\frac{\text d Y}{\text d X}=X\frac{\text d V}{\text d X}+V\]
\[X\frac{\text d V}{\text d X}=-\frac{5+2V}{2+V}-V\]\[X\frac{\text d V}{\text d X}=-\frac{5+2V+V(2+V)}{2+V}\]\[X\frac{\text d V}{\text d X}=-\frac{5+4V+V^2}{2+V}\]\[\frac 1X\text dX=-\frac{2+V}{5+4V+V^2}\text d V\]

Answer 4

\[\int\frac 1X\text dX=-\int \frac{V+2}{V^2+4V+5}\text d V\]\[\ln X=-\int \frac{V+2}{V^2+4V+5}\text d V\]\[\ln X=-\frac12\int \frac{2V+4}{V^2+4V+5}\text d V\]\[\ln X=-\frac12\ln(V^2+4V+5)+c\]\[X=e^c(V^2+4V+5)^{-1/2}\]
\[X(V^2+4V+5)^{1/2}=A\]\[X^2(V^2+4V+5)=A^2\]\[X^2\left(\frac{Y}{X}\right)^2+4X^2\left(\frac{Y}{X}\right)+5X^2=B\]\[Y^2+4XY+5X=B\]\[(y+3)^2+4(x-1)(y+3)+5(x-1)=B\]\[(y^2+6y+4)+4(xy+3x-y-3)+(5x-5)=B\]\[y^2+6y+9+4xy+12x-4y-12+5x-5=B\]\[y^2+4xy+17x+2y+4=B\]

Answer 5

have i done this right?

Answer 6

how can i check?

Answer 7

If you differentiate this
\[\frac{d}{dx}(y^2+4xy+17x+2y+4)=\frac{d}{dx}B\]
\[2y \frac{dy}{dx}+4y+4x\frac{dy}{dx}+17+2\frac{dy}{dx}=0\]
\[(2y+4x+2)\frac{dy}{dx}+4y+17=0\]
\[(y+2x+1)dy+(4y+17)dx=0\]
We should get back the original DE, you should check again!!

Answer 8

ok so i have got mistakes then

Answer 9

i dont follow your use of p and q
how can you get rid of the 1's

\[\frac{5x+2y+1}{2x+y+1} \neq \frac{5x+2y}{2x+y}\]
?

Answer 10

i have soved for p and q to get rid of the +1's

Answer 11

a technique my book calls
Homogeneous Equations Requiring a Change of Variables (LINEAR SHIFT)

Answer 12

Yeah that's right he can solve like that

Answer 13

@UnkleRhaukus thanks for showing how to solve this type of differential equations

Answer 14

yeah but i have it wrong somewhere

Answer 15

Effectively the \[x\rightarrow X+p\]\[y\rightarrow Y+p\]change of variables , simplifies the equation by centering it at the origin

Answer 16

\[\frac{\text d y}{\text dx}=\frac{\text d Y}{\text dX}\]

Answer 17

oh gotcha, didn't notice the change in variable...thanks

Answer 18

hmm

Answer 19

i found your mistake:
\[X^{2}(\frac{Y}{X})^{2}+4X^{2}(\frac{Y}{X})+5X^{2}=B\]
\[\rightarrow Y^{2}+4XY+5X = B\]
you forgot to square the X

Answer 20

\oh yeah thanks a bunch

Answer 21

\[X^2\left(\frac{Y}{X}\right)^2+4X^2\left(\frac{Y}{X}\right)+5X^2=B\]\[Y^2+4XY+5X^2=B\]\[(y+3)^2+4(x-1)(y+3)+5(x-1)^2=B\]\[(y^2+6y+4)+4(xy+3x-y-3)+5(x^2-2x+1)=B\]\[y^2+6y+9+4xy+12x-4y-12+5x^2-10x+5=B\]\[y^2+4xy+2x+2y+14=B\]

Answer 22

\[\frac{\text d }{\text dx}(y^2+4xy+2x+2y+14)=\frac{\text d }{\text dx}B\]
\[2y\frac{\text dy }{\text dx}+4y+4x\frac{\text d y}{\text dx}+2+2\frac{\text dy }{\text dx}=0\]
\[(2y+4x+2)\frac{\text dy }{\text dx}+4y+2=0\]

\[(2y+4x+2)\frac{\text dy }{\text dx}+=-(4y+2)\]
\[(2y+4x+2) \text dy =-(4y+2)\text d x\]
\[(2y+4x+2) \text dy +(4y+2)\text d x=0\]
\[(y+2x+1) \text dy +(2y+1)\text d x=0\]

Answer 23

almost right

Answer 24

just missing 5x

Answer 25

i got there