Q.Find a particular solution of the differential equation $(x - y)(dx + dy) = dx - dy$, given that $y = -1$, when $x = 0$. (Hint: put $x - y = t$)

Use the hint $t=x-y$, so $dt=dx-dy$ and (from $y=x-t$) $dy=dx-dt$, giving $dx+dy=2\,dx-dt$.

Substitute into $(x-y)(dx+dy)=dx-dy$:

Separate and integrate:

$2\,dx=\left(1+\frac{1}{t}\right)dt\ \Rightarrow\ 2x=t+\log|t|+C.$ …

Why the substitution helps

The combination $x-y$ appears on the left and $dx-dy$ on the right. Setting $t=x-y$ makes $dt=dx-dy$ appear directly, collapsing two variables into one.

Rewrite everything in $t$ and $x$

From $t=x-y$ we get $y=x-t$, so $dy=dx-dt$ and therefore

The equation $(x-y)(dx+dy)=dx-dy$ becomes

Separate

Expand and collect the $dt$ terms:

Divide by $t$ (the case $t=0$, i.e. $x=y$, fails the initial condition and is excluded):

Integrate

Restore $t=x-y$: …

Method: A hinted substitution to reduce to variable-separable

When a problem supplies a substitution like $x-y=t$, use it to collapse the equation to a separable one.

Steps

Step 1: Apply the given substitution.

Let $t=x-y$, so $dt=dx-dy$; express $dy$ (or $dx$) and any combination like $dx+dy$ in terms of $dx$ and $dt$.

Step 2: Rewrite the whole equation in $t$ and $x$.

Substitute so only $t$ and $x$ (and their differentials) remain. …

Common Mistakes

Mistake 1: Not using the given hint $x-y=t$.

Why it's wrong: the substitution is what collapses the equation; ignoring it leaves a tangled form. Correct approach: set $t=x-y$, so $dt=dx-dy$.

Mistake 2: Expressing $dx+dy$ incorrectly in terms of $t$.

Why it's wrong: from $y=x-t$, $dy=dx-dt$, so $dx+dy=2\,dx-dt$; a wrong expression derails the algebra. Correct approach: substitute $dy=dx-dt$ carefully. …