Determine the equation of the curve passing through the origin, in the form `y=f(x),` which satisfies the differential equation `(dy)/(dx)=sin(10 x+6y)dot`

Answer

`1/3 tan ^(-1) [4/5tan(4x+tan^(-1)frac{3}{4})-3/5]-(5x)/3`

Explanation

Ginve, `dy/dx=sin(10x+6y)`
Let `10x+6y=t …(i)`
`rArr 10+6dy/dx=(dt/dx)`
`rArr dy/dx=1/6(dt/dx-10)`
Now, the given differential equation becomes
`sin t = 1/6 (dt/dx -10)`
`rArr 6sin t = dt/dx -10`
`rArr dt/dx=6sin t +10`
`rArr dt/(6sin t +10)=dx`
On integrating both sides, we get
`1/2intdt/(3sint+5)=x+c …(ii)`
Let `I_(1)=intdt/(3sint+5)=intdt/(3((2tan t//1)/(1+tan^(2)t//2))+5)`
`=int ((1+tan^(2) t//2)dt)/((6 tan frac{t}{2} +5+5tan^(2)frac{t}{2})` ltbr. Put `tan t//2=u`
`rArr 1/2 sec^(2) t//2 dt =du rArr dt= (2du)/(sec^(2) t//2)`
`rArr dt=(2du)/(1+tan^(2)t//2)rArr dt = (2du)/(1+u^(2))`
`therefore I_(1)=int (2(1+u^(2))du)/((1+u^(2))(5u^(2)+6u+5))=2/5 int (du)/(u^(2)+6/5u+1)`
`=2/5 int (du)/(u^(2)+6/5u+9/25-9/25+1)`
`=2/5int(du)/((u+3/5)^(2)+(4/5)^(2))=2/5 cdot 5/4 tan^(-1) ((u+3//5)/(4//5))`
`= 1/2 tan ^(-1)[(5u=3)/4]=1/2 tan ^(-1) [(5tan t//2+3)/4]`
on putting this in Eq. (ii), we get
`1/4 tan^(-1) [(5tanfrac {t}{2}+3)/4]=x+c`
` rArr tan^(-1) [(5tanfrac {t}{2}+3)/4]=4x+4c`
`rArr 1/4[5tan (5x+3y)+3] =tan (4x+4c)`
`rArr 5tan (5x+3y)+3 =4tan (4x+4c)`
When `x=0, y=0,` we get
`5 tan 0 + 3=4 tan (4c)`
`rArr 3/4= tan 4c`
`rArr 4c= tan ^(-1)frac{3}{4}`
Then, `5tan (5x+3y)+3=4tan (4x+tan^(_1) frac{3}{4})`
`rArr tan (5x+3y)=4/5tan (4x+tan frac{3}{4})-3/5`
`rArr 5x+3y=tan^(-1)[4/5{tan(4x+tan^(-1) frac{3}{4})}-3/5]`
`rArr 3y=tan^(-1)[4/5{tan(4x+tan^(-1) frac{3}{4})}-3/5]-5x`
`rArr y=1/3tan^(-1)[4/5{tan(4x+tan^(-1) frac{3}{4})}-3/5]-(5x)/3`