We're looking for an integrating factor [tex]\mu(x,y)=x^py^q[/tex] such that
[tex]\mu\underbrace{(49y^3+45xy)}_M\,\mathrm dx+\mu\underbrace{(98xy^2+50x^2)}_N\,\mathrm dy=0[/tex]
is exact, which would require that
[tex](\mu M)_y=(\mu N)_x[/tex]
[tex](49x^py^{q+3}+45x^{p+1}y^{q+1})_y=(98x^{p+1}y^{q+2}+50x^{p+2}y^q)_x[/tex]
[tex]49(q+3)x^py^{q+2}+45(q+1)x^{p+1}y^q=98(p+1)x^py^{q+2}+50(p+2)x^{p+1}y^q[/tex]
[tex]\implies\begin{cases}49(q+3)=98(p+1)\\45(q+1)=50(p+2)\end{cases}\implies p=\dfrac52,q=4[/tex]
You can verify that [tex](\mu M)_y=(\mu N)_x[/tex] if you'd like. With the ODE now exact, we have a solution [tex]F(x,y)=C[/tex] such that
[tex]F_x=\mu M[/tex]
[tex]F=\displaystyle\int(49y^3+45xy)x^{5/2}y^4\,\mathrm dx[/tex]
[tex]F=10x^{9/2}y^5+14x^{7/2}y^7+f(y)[/tex]
[tex]F_y=\mu N[/tex]
[tex]50x^{9/2}y^4+98x^{7/2}y^6+f'(y)=98x^{7/2}y^2+50x^{9/2}y^4[/tex]
[tex]f'(y)=0[/tex]
[tex]\implies f(y)=C[/tex]
and so the general solution is
[tex]F(x,y)=10x^{9/2}y^5+14x^{7/2}y^7=C[/tex]
